Contoh Soal 6 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{ll}\\ 26.&\textrm{Misalkan}\: \: F=\displaystyle \frac{6x^{2}+16x+3m}{6}\: \: \textrm{merupakan kuadrat}\\ &\textrm{dari bentuk linear terhadap}\: \: x(ax+b).\: \: \textrm{Nilai}\\ &m\: \: \textrm{yang memungkinkan kondisi tersebut terjadi}\\ &\textrm{terletak di antara}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -6\: \: \textrm{dan}\: \: -5&&&&&\textrm{D}.&\displaystyle \textrm{4 dan 5}\\ \textrm{B}.&\displaystyle -4\: \: \textrm{dan}\: \: -3&&&\displaystyle &&\textrm{E}.&\displaystyle \textrm{5 dan 6}\\ \textrm{C}.&\color{red}\displaystyle 3\: \: \textrm{dan}\: \: 4 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:\displaystyle \frac{6x^{2}+16x+3m}{6}\\ &\textrm{merupakan bentuk kuadrat, maka}\\ &f(x)=\displaystyle \frac{6x^{2}+16x+3m}{6}=x^{2}+\displaystyle \frac{8}{3}x+\frac{1}{2}m\\ &\textrm{dengan}\: \: a=1,\: b=\displaystyle \frac{8}{3},\: \: \textrm{dan}\: \: \: c=\displaystyle \frac{1}{2}m\\ &\textrm{dan istilah bentuk linier mengarah kepada}\\ &\textrm{jenis akarnya real, maka}\: \: D=b^{2}-4ac>0\\ &\left ( \displaystyle \frac{8}{3} \right )^{2}-4.1.\left ( \displaystyle \frac{1}{2}m \right )>0\\ &\Leftrightarrow \displaystyle \frac{64}{9}-2m>0\\ &n \to \Leftrightarrow \displaystyle \frac{32}{9}-m>0\: \: \: (\textrm{dikali})-1\\ &m-\displaystyle \frac{32}{9}>0\\ &\Leftrightarrow m>\displaystyle \frac{32}{9}\Leftrightarrow m>\color{red}3\displaystyle \frac{5}{9} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Ekspresi}\: \: 21x^{2}+ax+21\: \: \textrm{dapat dituliskan}\\ &\textrm{sebagai perkalian dua bentuk linear dengan}\\ &\textrm{koefisien bilangan bulat. Hal tersebut hanya}\\ &\textrm{dapat dilakukan jika}\: \: a\: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle \textrm{bilangan 0}&&&&&\\ \textrm{B}.&\displaystyle \textrm{beberapa bilangan ganjil}&&\\ \textrm{C}.&\color{red}\textrm{beberapa bilangan genap}\\ \textrm{D}.&\textrm{sembarang bilangan ganjil}\\ \textrm{E}.&\textrm{sembarang bilangan genap} \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\textrm{Ekspresi dapat dituliskan dalam perkalian dua}\\ &\textrm{bentuk linear, maka}\: D=b^{2}-4ac\geq 0\\ &\textrm{Dari}\: \: 21x^{2}+ax+21\\&a=21,\: b=a,\: c=21,\: \: \textrm{maka}\\ &a^{2}-4.21.21\geq 0\\ &\Leftrightarrow a^{2}-42^{2}\geq 0\\ &\Leftrightarrow (a+42)(a-42)\geq 0\\ &\Leftrightarrow \color{red}a\leq -42\: \: \color{black}atau\: \: \color{red}a\geq 42 \end{array}$.

Hubungan Fungsi Kuadrat dan Sebuah garis linear

$\begin{array}{ll}\\ 28.&\textrm{Garis}\: \: 2x+y=p\: \: \textrm{akan memotong parabola}\\ &4x^{2}-y=0\: \: \textrm{di dua titik apabila nilai}\: \: p=\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&p<-\displaystyle \frac{1}{4}&&&&&\textrm{D}.&\displaystyle p<-\frac{3}{4}\\ \textrm{B}.&\color{red}\displaystyle p>-\displaystyle \frac{1}{4}&&&\displaystyle &&\textrm{E}.&\displaystyle p<-1\\ \textrm{C}.&\displaystyle p<\displaystyle \frac{1}{4}\end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui bahwa garis}\: \: 2x+y=p\Rightarrow y=p-2x\\ &\textrm{memotong parabola}\: \: 4x^{2}-y=0\: \:  \textrm{di dua titik}\\ &\textrm{berbeda, maka}\\ &4x^{2}-y=0\Leftrightarrow 4x^{2}-(p-2x)=0\\ &4x^{2}+2x-p=0,\: \: \textrm{dengan}\: a=4,b=2,\: c=-p\\ &\textrm{Syarat memotong di dua titik berbeda adalah}:\\ &D=b^{2}-4ac>0\\ &\Leftrightarrow 2^{2}-4.4.(-p)>0\\ &\Leftrightarrow 4+16p>0\Leftrightarrow 16p>-4\Leftrightarrow p>-\displaystyle \frac{4}{16}\\ &\Leftrightarrow p>\color{red}-\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: ax^{2}+bx+c=0\: \: \textrm{tidak mempunyai akar real}\\ &\textrm{maka grafik fungsi}\: \: y=ax^{2}+bx+c\: \: \textrm{akan}\\ &\textrm{menyinggung garis}\: \: y=x\: \: \textrm{apabila}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}b<\displaystyle \frac{1}{2}&&&&&\textrm{D}.&\displaystyle b>2\\ \textrm{B}.&\displaystyle b>\displaystyle \frac{1}{2}&&&\displaystyle &&\textrm{E}.&\displaystyle 1<b<2\\ \textrm{C}.&\displaystyle b>1\end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui sebuah kurva parabola}\\ &y=ax^{2}+bx+c\: \: (\textrm{tidak punya akar real})\\ &\textrm{dan}\: \: \textrm{garis}\: \: y=x\: \: \textrm{saling bersinggungan, }\\&\textrm{sehingga}\: \: y=y\\ &\Leftrightarrow ax^{2}+bx+c=x\\ &\Leftrightarrow ax^{2}+bx-x+c=0\\ &\Leftrightarrow ax^{2}+(b-1)x+c=0\\ &\textrm{dengan}\: \: a=a,\: b=b-1,\: c=c \\&\textrm{Selanjutnya syarat menyinggung adalah}:\\ &D=b^{2}-4ac=0\\ &\Leftrightarrow (b-1)^{2}-4ac=0\Leftrightarrow (b-1)^{2}=4ac.\\ &\textrm{Karena}\: \: ax^{2}+bx+c\: \: (\textrm{tidak punya akar real})\\ &\textrm{maka}\: \: D<0\Leftrightarrow b^{2}-4ac<0\\ &\Leftrightarrow b^{2}-(b-1)^{2}<0\Leftrightarrow b^{2}-(b^{2}-2b+1)<0\\ &\Leftrightarrow 2b-1<0\Leftrightarrow b<\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Grafik}\: \: y=2x^{2}+ax+b\: \: \textrm{berpotongan dengan}\\ &\textrm{garis}\: \: y=3x-1\: \: \textrm{di titik}\: \: (x_{1},y_{1})\: \: \textrm{dan}\: \: (x_{2},y_{2})\\ &\textrm{Jika}\: \: x_{1}+x_{2}=3\: \: \textrm{dan}\: \: x_{1}\times x_{2}=4,\: \textrm{maka}\\ &a+b=....\\ &\begin{array}{lllllllll}\textrm{A}.&1&&&&&\textrm{D}.&\color{red}\displaystyle 4\\ \textrm{B}.&\displaystyle 2&&&\displaystyle &&\textrm{E}.&\displaystyle 5\\ \textrm{C}.&\displaystyle 3\end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ & \end{array}$

Contoh Soal 5 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: y=\sqrt{x^{2}-2x+1}+\sqrt{x^{2}+2x+1}\\ &\textrm{maka nilai}\: \: y=....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle \displaystyle 2x&&&&&\textrm{D}.&\color{red}\displaystyle \left | x-1 \right |+\left | x+1 \right |\\ \textrm{B}.&\displaystyle 2(x+1)&&\textrm{C}.&\displaystyle 0&&\textrm{E}.&\displaystyle \textrm{tidak ada yang benar} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui}\: \: y=\sqrt{x^{2}-2x+1}+\sqrt{x^{2}+2x+1}\\ &\Leftrightarrow y=\sqrt{(x-1)^{2}}+\sqrt{(x+1)^{2}}\\ &\Leftrightarrow y=\color{red}\left | x-1 \right |+\left | x+1 \right | \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: x\: \: \textrm{bilangan real dan}\: \: 4y^{2}+4xy+x+6=0,\\ &\textrm{maka nilai semua nilai}\: \: x\: \: \textrm{agar nilai}\: \: y\: \: \textrm{real adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle x\leq -2\: \: \textrm{atau}\: \: x\geq 3&&&&&\textrm{D}.&\displaystyle -3\leq x\leq 2\\ \textrm{B}.&\displaystyle x\leq 2\: \: \textrm{atau}\: \: x\geq 3&&&\displaystyle &&\textrm{E}.&\displaystyle -2\leq x\leq 3\\ \textrm{C}.&\displaystyle x\leq -3\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&4y^{2}+(4x)y+(x+6)=0\\ &\textrm{agar}\: \: y\: \: \textrm{real, maka}\: \: D=b^{2}-4ac\geq 0\\ &(4x)^{2}-4(4)(x+6)\geq 0\\ &\Leftrightarrow 16x^{2}-16(x+6)\geq 0\\ &\Leftrightarrow x^{2}-x-6\geq 0\\ &\Leftrightarrow (x-3)(x+2)\geq 0\\ &\Leftrightarrow \color{red}\displaystyle x\leq -2\: \: \textrm{atau}\: \: x\geq 3 \end{aligned}\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\  \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle 2&&&&3&&\textbf{Sumbu-X}\\  \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}x=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{x(x-1)}{2}\: ,\: \textrm{nilai}\: \: f(x+2)= ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle f(x)+f(2)&&&&&\textrm{D}.&\displaystyle \frac{xf(x)}{x+2}\\ \textrm{B}.&\displaystyle (x+2)f(x)&&&\displaystyle &&\textrm{E}.&\color{red}\displaystyle \displaystyle \frac{(x+2)f(x+1)}{x}\\ \textrm{C}.&\displaystyle x(x+2)f(x) \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &f(x)=\displaystyle \frac{x(x-1)}{2},\\ &f(x+1)=\displaystyle \frac{(x+1)(x)}{2}\Leftrightarrow \displaystyle \frac{x+1}{2}=\frac{f(x+1)}{x}\: ,\: \textrm{maka}\\ &\begin{aligned}f(x+2)&=\displaystyle \frac{(x+2)((x+2)-1)}{2}=\frac{(x+2)(x+1)}{2}\\ &=\displaystyle \frac{f(x+1)}{x}.(x+2)=\color{red}\displaystyle \displaystyle \frac{(x+2)f(x+1)}{x} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Jika}\: \: r_{1}\: \: \textrm{dan} \: \: r_{2}\: \: \textrm{merupakan dua penyelesaian }\\ &\textrm{dari persamaan}\: \: x^{2}+px+8=0\: ,\: \textrm{maka}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \left | r_{1}+r_{2} \right |>4\sqrt{2}&&&\textrm{D}.&\displaystyle r_{1}<0\: \: \textrm{dan}\: \: r_{2}<0\\ \textrm{B}.&\displaystyle \left | r_{1} \right |>3\: \: \textrm{dan}\: \: \left | r_{2} \right |>3 &\displaystyle &&\textrm{E}.&\displaystyle -\left | r_{1}+r_{2} \right |<4\sqrt{2}\\ \textrm{C}.&\displaystyle \left | r_{1} \right |>2\: \: \textrm{dan}\: \: \left | r_{2} \right |>2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui PK}:x^{2}+px+8=0\\ &\textrm{dengan}\left\{\begin{matrix} r_{1}\\  r_{2} \end{matrix}\right.\: \: \textrm{akar-akarnya, maka}\\ &\bullet \quad r_{1}+r_{2}=-p\\ &\bullet \quad r_{1}\times r_{2}=8\\ &\textrm{Asumsikan kedua akarnya real dan berbeda,}\\&\textrm{sehingga kita pilih nilai} \: \: \: D=b^{2}-4ac>0\\  &\Leftrightarrow (-p)^{2}-4.1.8>0\\ &\Leftrightarrow p^{2}-32>0\\ &\Leftrightarrow (\left | r_{1}+r_{2} \right |+\sqrt{32})(\left | r_{1}+r_{2} \right |-\sqrt{32})>0\\ &\Leftrightarrow \left | r_{1}+r_{2} \right |<-\sqrt{32}\: \: \textrm{atau}\: \: \color{red}\left | r_{1}+r_{2} \right |>\sqrt{32}\\&\qquad\overset{\begin{matrix} \Downarrow\\  \Downarrow \end{matrix}}{\textbf{tidak mungkin}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: (1,y_{1})\: \: \textrm{dan}\: \: (-1,y_{2})\: \: \textrm{terletak pada}\\ &\textrm{grafik}\: \: y=ax^{2}+bx+c\: \: \textrm{dan}\: \: y_{1}-y_{2}=-6\\ &\textrm{maka nilai}\: \: b=....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \displaystyle -3&&&&&\textrm{D}.&\displaystyle \sqrt{ac}\\ \textrm{B}.&\displaystyle 0&&\textrm{C}.&\displaystyle 3&&\textrm{E}.&\displaystyle \displaystyle \frac{a+c}{2} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui}\: \: y=ax^{2}+bx+c,\: \: \textrm{untuk}\\ &\bullet \quad (1,y_{1})\Rightarrow y_{1}=a+b+c\\ &\bullet \quad (-1,y_{1})\Rightarrow y_{2}=a-b+c\\ &\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ -\\ &\qquad\qquad y_{1}-y_{2}= 2b=-6\Rightarrow b=\color{red}-3\end{array}$.

Contoh Soal 4 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 16.&\textrm{Batas nilai}\: \: m\: \: \textrm{yang memenuhi agar fungsi}\\ &\textrm{kuadrat}\: \: y=mx^{2}+(m+2)x+m\: \: \\ &\textrm{memotong sumbu-X di dua titik yang}\\ &\textrm{berbeda adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle -\displaystyle \frac{2}{3}<m<2&&&&&\\ \textrm{B}.&\displaystyle -2<m<\displaystyle \frac{2}{3}&&\\ \textrm{C}.&m<-2\: \: \textrm{atau}\: \: m>2\\ \textrm{D}.&m<-2\: \: \textrm{atau}\: \: m>\displaystyle \frac{2}{3}\\ \textrm{E}.&m<-\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: m>2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: y=mx^{2}+(m+2)x+m\\ &\textrm{dengan}\: \: a=m,\: b=m+2,\: \&\: \: c=m\\ &\textrm{memotong sumbu-X di dua titik berbeda}\\ &\textrm{hal ini artinya nilai Diskriminan }D>0\\ &D=b^{2}-4ac>0\\ &(m+2)^{2}-4m.m>0\\ &\Leftrightarrow m^{2}+4m+4-4m^{2}>0\\ &\Leftrightarrow -3m^{2}+4m+4>0\: \: \: (\textrm{dikali} -1)\\ &\Leftrightarrow 3x^{2}-4m-4<0\\ &\Leftrightarrow (m-2)(3m+2)\color{red}<0\\ &\Leftrightarrow \color{red}-\displaystyle \frac{2}{3}<m<2\\ &\textbf{Anda bisa menggunakan titik uji dulu}\\ &\textrm{untuk memastikannya wilayah yg dimaksud}\\ &\textrm{misalkan pilih}\: \: m=0\\ &\textrm{lalu kita ujikan, yaitu}:\\ &m=0\Rightarrow (0-2)(3.0+2)=-4\color{red}<0\: \: (\textbf{benar})\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\  \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle \frac{2}{3}&&&&2&&\textbf{Sumbu}-m\\  \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}m=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Batas nilai}\: \: m\: \: \textrm{yang menyebabkan fungsi kuadrat}\\ & y=(m-1)x^{2}-2(m-1)x+(2m+1)\: \: \\ &\textrm{definit positif adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle m>-2&&&&&\\ \textrm{B}.&\color{red}\displaystyle m>1&&\\ \textrm{C}.&m<1\\ \textrm{D}.&-2<m<1\\ \textrm{E}.&m<-\displaystyle 2\: \: \textrm{atau}\: \: m>1 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:\\ &y=(m-1)x^{2}-2(m-1)x+(2m+1)\\ &\textrm{Syarat fungsi kuadrat definit positif}:\\ &\bullet\quad a>0\\ &\bullet \quad D=b^{2}-4ac<0\\ &\textrm{maka}\\ &a=m-1>0\Leftrightarrow m>1,\: \: \textrm{dan}\\ &D=(2(m-1))^{2}-4(m-1)(2m+1)<0\\ &\Leftrightarrow 4(m-1)^{2}-(m-1)(8m+4)<0\\ &\Leftrightarrow (m-1)(4m-4)-(m-1)(8m+4)<0\\ &\Leftrightarrow (m-1)(4m-4-8m-4)<0\\ &\Leftrightarrow (m-1)(-4m-8)<0\: ,\: \textbf{dibagi}\: \: -4\\ &\Leftrightarrow (m-1)(m+2)\color{red}>0 \end{aligned}\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\  \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle 2&&&&1&&\textbf{Sumbu}-m\\  \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}m=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Luas}\: \: L\: \: \textrm{suatu segitiga}\: \: ABC\: \: \textrm{diketahui}\\ & x(7-x)\: \: \textrm{cm}^{2}.\: \textrm{Luas maksimum segitiga} \\ &\textrm{tersebut adalah}\: \: ...\: \: \textrm{cm}^{2}\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 3\displaystyle \frac{1}{2}&&&&&\textrm{D}.&\displaystyle 10\displaystyle \frac{1}{4}\\\\ \textrm{B}.&\displaystyle 5\displaystyle \frac{1}{2}&&\textrm{C}.&\displaystyle 7\displaystyle \frac{1}{4}&&\textrm{E}.&\color{red}\displaystyle 12\displaystyle \frac{1}{4} \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &\textrm{Diketahui}\: \: \left [ ABC \right ]=x(7-x)=7x-x^{2}\\ &\textrm{dengan}\: \: a=-1,\: \: b=7,\: \textrm{dan}\: \: c=0\\ &\textrm{akan}\: \textbf{maksimum},\: \textrm{saat}\: \: \left (x_{ss},f(x_{ss})\right )\\ &\textrm{yaitu}:\\ &x_{ss}=-\displaystyle \frac{b}{2a}=-\frac{7}{2.(-1)}=\frac{7}{2},\: \: \textrm{maka}\\ &f\left ( \displaystyle \frac{7}{2} \right )=7\left ( \displaystyle \frac{7}{2} \right )-\left ( \displaystyle \frac{7}{2} \right )^{2}=\displaystyle \frac{49}{2}-\frac{49}{4}=\color{red}\displaystyle \frac{49}{4} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Perhatikan gambar persegi ABCD dengan}\\ &\textrm{panjang sisinya  10 cm} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Jika}\: \: BP=DQ=x\: \: \textrm{cm, maka luas }\\ &\textrm{maksimum segitiga}\: \: APQ\: \: \textrm{adalah}\: ...\: \: \textrm{cm}^{2}\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 35&&&&&\textrm{D}.&\displaystyle 75\\ \textrm{B}.&\color{red}\displaystyle 50&&\textrm{C}.&\displaystyle 60&&\textrm{E}.&80 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}\left [ APQ \right ]&=L_{\square ABCD}-\left [ ADQ \right ]-\left [ QCP \right ]-\left [ APB \right ]\\ &=10^{2}-\displaystyle \frac{1}{2}x.10-\frac{1}{2}.(10-x)^{2}-\frac{1}{2}x.10\\ &=100-10x-\displaystyle \frac{1}{2}(10-x)^{2}\\ &=10(10-x)-\displaystyle \frac{1}{2}(10-x)^{2}\\ &=(10-x)\left ( 10-\displaystyle \frac{1}{2}(10-x) \right )\\ &=(10-x)\left ( 5+\displaystyle \frac{1}{2}x \right )\\ &=50-\displaystyle \frac{1}{2}x^{2} \end{aligned}\\ &x_{ss}=-\displaystyle \frac{b}{2a}=-\displaystyle \frac{0}{2.\displaystyle \frac{1}{2}}=0,\: \: \textrm{maka nilai}\\ &f(x_{ss})=f(0)=50-\displaystyle \frac{1}{2}.0^{2}=\color{red}50 \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Fungsi}\: \: f(x)=a(b-c)x^{2}+b(c-a)x+c(a-b)\\ &\textrm{akan memotong sumbu-X di titik}\: \: (1,0)\: \: \textrm{dan}\\ &\textrm{memenuhi}\: \: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle \displaystyle \frac{b(c-a)}{a(b-c)}&&&&&\textrm{D}.&\color{red}\displaystyle \displaystyle \frac{c(a-b)}{a(b-c)}\\\\ \textrm{B}.&\displaystyle \displaystyle \frac{a(b-c)}{c(a-b)}&&\textrm{C}.&\displaystyle \displaystyle \frac{a(b-c)}{b(c-a)}&&\textrm{E}.&\displaystyle \displaystyle \frac{c(a-b)}{b(c-a)} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui bahwa FK}:\\ &f(x)=a(b-c)x^{2}+b(c-a)x+c(a-b)\\ &\textrm{maka nilai}\\ &\bullet \quad x_{1}+x_{2}=-\displaystyle \frac{b(c-a)}{a(b-c)}=\displaystyle \frac{b(a-c)}{a(b-c)}\\ &\bullet \quad x_{1}\times x_{2}=\color{red}\displaystyle \frac{c(a-b)}{a(b-c)} \end{array}$

Contoh Soal 3 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{ll}\\ 11.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan grafik fungsi kuadrat pada}\\ &\textrm{pada gambar tersebut di atas dengan}\\ &\textrm{titik balik}\: \: \left ( -2,\displaystyle \frac{9}{2} \right )\: \: \textrm{dan melalui}\: (-1,4)\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y=\displaystyle \frac{1}{2}\left (2+4x-x^{2}  \right )&&&&&\\ \textrm{B}.&\color{red}\displaystyle y=\displaystyle \frac{1}{2}\left (5-4x-x^{2}  \right )&&\\ \textrm{C}.&y=\displaystyle \frac{1}{2}\left (5-2x-x^{2}  \right )\\ \textrm{D}.&y=1-4x-x^{2}\\ \textrm{E}.&y=5+3x-x^{2} \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}+bx+c,\: \: \textrm{dengan}\\ &\textrm{koordinat}\: \: \left ( x_{ss},y_{ss} \right )=\left ( -2,\displaystyle \frac{9}{2} \right )\: \: \textrm{dan melalui}\\ &\textrm{titik}\: \: (-1,4)\: ,\: (\color{red}-5\color{black},0)\: ,\: \textrm{serta}\: \: (\color{red}1\color{black},0),\: \textrm{maka}\\ &y=a(x-x_{ss})^{2}+y_{ss}=a(x-x_{1})(x-x_{2})\\ &\textrm{Pilih salah satunya, di sini saya pilih formula}\\ &\textrm{yang kedua, yaitu}:\color{red}y=a(x-x_{1})(x-x_{2})\\ &\Leftrightarrow 4=a(-1-(-5))(-1-1)\Leftrightarrow 4=a(4)(-2)\\ &\Leftrightarrow a=-\displaystyle \frac{1}{2}\\ &\textrm{Selanjutnya kembalikan ke formula semula}\\ &\textrm{yaitu}:\\ &y=a(x-x_{1})(x-x_{2})=-\displaystyle \frac{1}{2}(x-(-5))(x-1)\\ &\Leftrightarrow y=-\displaystyle \frac{1}{2}(x+5)(x-1)=\color{red}\displaystyle \frac{1}{2}(5-4x-x^{2}) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Gambar berikut adalah fungsi parabola}\\ &\textrm{dengan persamaan}\: \: y=ax^{2}-4x+k \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Jika nilai minimum}\: \: y\: \: \textrm{adalah}\: \: -8\: \: \textrm{maka}\\ &\textrm{nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -5&&&&&\textrm{D}.&\displaystyle 2\\ \textrm{B}.&\displaystyle -4&&\textrm{C}.&\color{red}\displaystyle -2&&\textrm{E}.&5 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}-4x+k,\: \: \textrm{dengan}\\ &\textrm{koordinat}\: \: \left ( x_{ss},y_{ss} \right )=\left ( 3,-8 \right )\: \: \textrm{maka}\\ &y=a(x-x_{ss})^{2}+y_{ss}\\ &-2=a(0-3)^{2}+(-8)\Leftrightarrow -2+8=9.a\\&6=9a\Leftrightarrow a=\displaystyle \frac{6}{9}=\frac{2}{3}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\displaystyle \frac{2}{3}(x-3)^{2}-8=\frac{2}{3}(x^{2}-6x+9)-8\\ &\Leftrightarrow y=\displaystyle \frac{2}{3}x^{2}-4x+6-8=\color{red}\displaystyle \frac{2}{3}x^{2}-4x-2\\ &\textrm{Jadi, nilai}\: \: \: k=\color{red}-2 \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 13.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan grafik fungsi kuadrat pada}\\ &\textrm{gambar di atas adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y=-(x-3)^{2}-1&&&&&\\ \textrm{B}.&\displaystyle y=-(x-3)^{2}+1&&\\ \textrm{C}.&y=-\displaystyle \frac{1}{3}\left (x-3 \right )^{2}+1\\ \textrm{D}.&\color{red}y=-\displaystyle \frac{1}{3}\left (x-3 \right )^{2}-1\\ \textrm{E}.&y=-\displaystyle \frac{1}{3}\left (x+3 \right )^{2}-1 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Diketahui FK dengan koordinat}\\ &\left ( x_{ss},y_{ss} \right )=\left ( 3,-1 \right )\: \: \textrm{dan melalui titik}\: \: (0,-4)\\ &\textrm{maka}\: \: y=a(x-x_{ss})^{2}+y_{ss}\\ &-4=a(0-3)^{2}+(-1)\Leftrightarrow -4+1=9.a\\&-3=9a\Leftrightarrow a=-\displaystyle \frac{1}{3}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\color{red}-\displaystyle \frac{1}{3}(x-3)^{2}-1 \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 14.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Jika grafik fungsi kuadrat di atas adalah}\\ &y=ax^{2}+bx+c,\: \textrm{maka hasil kali dari}\\ &a.b.c\: \: \textrm{ adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -20&&&&&\textrm{D}.&\displaystyle 3\\ \textrm{B}.&\displaystyle -6&&\textrm{C}.&\color{red}\displaystyle -3&&\textrm{E}.&20 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:ax^{2}+bx+c\\ &\left ( x_{ss},y_{ss} \right )=\left ( -2,-4 \right )\: \: \textrm{dan melalui titik}\: \: (0,-6)\\ &\textrm{maka}\: \: y=a(x-x_{ss})^{2}+y_{ss}\\ &-6=a(0-(-4))^{2}+(-2)\Leftrightarrow -6+2=16.a\\&-4=16a\Leftrightarrow a=-\displaystyle \frac{1}{4}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\color{red}-\displaystyle \frac{1}{4}(x+4)^{2}-2\\ &=-\displaystyle \frac{1}{4}x^{2}-2x-6\left\{\begin{matrix} a=-\displaystyle \frac{1}{4}\\  b=-2\\  c=-6 \end{matrix}\right.\\ &\textrm{Sehingga nilai}\: \: abc=\left ( -\displaystyle \frac{1}{4} \right ).(-2).(-6)=\color{red}-3 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Grafik fungsi kuadrat}\: \: y=f(x)=x^{2}\\ &\textrm{digeser 1 satuan ke kanan dan dilanjutkan}\\ &\textrm{1 satuan ke atas. Persamaan parabola }\\ &\textrm{yang baru adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y+1=(x+1)^{2}&&&&&\\ \textrm{B}.&\displaystyle y+1=x^{2}+1&&\\ \textrm{C}.&y-1=x^{2}-1\\ \textrm{D}.&\color{red}y-1=(x-1)^{2}\\ \textrm{E}.&y=(x+1)^{2}+1 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Sebagai pedoman bantuan, suatu fungsi}\\ &\textrm{di geser ke kanan berarti}:x-1\: \: \textrm{dan}\\ &\textrm{digeser ke atas berarti}:y-1\\ &\textbf{Catatan}:\\ &\textrm{Andai digeser ke kiri 1 kemudian ke bawah 1,}\\ &\textrm{maka garfik akan menjadi}:y+1=(x+1)^{2}\\&\textrm{Berikut ilustrasi grafiknya}  \end{array}$.

Contoh Soal 2 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{ll}\\ 6.&\textrm{Fungsi kuadrat}\: \: f(x)=(2x+p)^{2}+q\\ &\textrm{dengan titik balik minimum}\: \:  (-1,3).\\ &\textrm{Nilai}\: \: p+q\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 2&&&&&\textrm{D}.&\displaystyle 6\\ \textrm{B}.&\displaystyle 4&&\textrm{C}.&\color{red}\displaystyle 5&&\textrm{E}.&7 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=(2x+p)^{2}+q,\: \: \textrm{dengan}\\ &\textrm{titik}\: \: \left ( x_{ss},y_{ss} \right )=(-1,3),\: \: \textrm{maka}\\ &f(x)=4x^{2}+4px+p^{2}+q\: \: \textrm{dengan}\\ &x_{ss}=-1=-\displaystyle \frac{b}{2a}\Leftrightarrow 1=\displaystyle \frac{4p}{2.4}\Leftrightarrow p=2\\ &\textrm{Selanjutnya}\\ &f(-1)=(2.(-1)+2)^{2}+q=3\Leftrightarrow q=3\\ &\textrm{maka}\\ &p+q=\color{red}2+3=5 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika grafik fungsi kuadrat}\: \: f(x)=ax^{2}+x+c\\ &\textrm{dengan titik balik minimum}\: \:  (-1,3)\: \: \textrm{dan melalui}\\ &(2,12)\: \: \textrm{maka}\: \: a+b+c\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle 7&&&&&\textrm{D}.&\displaystyle 13\\ \textrm{B}.&\displaystyle 9&&\textrm{C}.&\displaystyle 11&&\textrm{E}.&15 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\&\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}+bx+c,\: \: \textrm{dengan}\\ &\textrm{titik}\: \: \left ( x_{ss},y_{ss} \right )=(-1,3)\: \: \textrm{dan melalui titik}\\ &(2,12)\: ,\: \textrm{maka}\\ &\begin{array}{c|c}\hline \begin{aligned} &12=4a+2b+c\\ &3=a-b+c\\&\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ -\\ &9=3a+3b\\ &\Leftrightarrow 3=a+b \end{aligned}&\begin{aligned}&-\displaystyle \frac{b}{2a}=-1,\: \: \textrm{maka}\\&2a-b=0,\: \: \textrm{dan ingat}\\ &a+b=3\\ &\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ +\\ &3a=3\\ &\Leftrightarrow a=1,\: \: \textrm{maka}\: \: b=2 \end{aligned} \end{array}\\ &\textrm{dan}\\ &a-b+c=3\Rightarrow 1-2+c=3\Rightarrow c=4\\ &\textrm{Jadi, nilai}\: \: a+b+c=\color{red}1+2+4=7  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai minimum grafik fungsi}\: \: f(x)=ax^{2}-2x+8\\ &\textrm{adalah 5. Nilai }\: \:  6a\: \: \: \textrm{sama dengan}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 1&&&&&\textrm{D}.&\displaystyle 9\\ \textrm{B}.&\color{red}\displaystyle 2&&\textrm{C}.&\displaystyle 4&&\textrm{E}.&12 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}-2x+8,\: \: \textrm{dengan}\\ &\textrm{titik}\: \: \color{red}\left ( x_{ss},y_{ss} \right )=\left (-\displaystyle \frac{b}{2a},5  \right )\: \: \color{black}\textrm{maka}\\ &\bullet \quad x_{ss}=-\displaystyle \frac{b}{2a}=-\frac{-2}{2a}=\color{red}\frac{1}{a}\\ &\bullet \quad y_{ss}=f\left ( x_{ss} \right )=a\left ( \displaystyle \frac{1}{a} \right )^{2}-2\left ( \displaystyle \frac{1}{a} \right )+8=\color{red}5\\ &\qquad\quad \Leftrightarrow \displaystyle \frac{1}{a}-\frac{2}{a}=5-8\Leftrightarrow -\displaystyle \frac{1}{a}=-3\\ &\qquad\quad \Leftrightarrow a=\displaystyle \frac{1}{3}\\ &\textrm{maka nilai}\: \: 6a=6\left ( \displaystyle \frac{1}{3} \right )=\color{red}2  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika kurva fungsi}\: \: f(x)=x^{2}+bx+c\\ &\textrm{memotong sumbu-X di}\: \:  (1,0)\: \: \: \textrm{dan}\: \: (5,0),\\ &\textrm{maka nilai}\: \: b^{2}-c^{2}\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -11&&&&&\textrm{D}.&\color{red}\displaystyle 11\\ \textrm{B}.&\displaystyle -3&&\textrm{C}.&\displaystyle 6&&\textrm{E}.&13 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=x^{2}+bx+c,\: \: \textrm{memotong}\\ &\textrm{sumbu-X di}\: \: (1,0)\:  \& \: (5,0)\: \: \textrm{artinya}\: x_{1}=1\: \&\: x_{2}=5\\ &\textrm{maka}\\ &x_{ss}=\displaystyle \frac{-b}{2.1}=\displaystyle \frac{x_{1}+x_{2}}{2}=\frac{1+5}{2}\Leftrightarrow b=-6\\ &\textrm{dan kita juga memiliki}\\ &f(1)=1+b+c=0\Rightarrow c=-b-1=6-1=5\\ &\textrm{Sehingga}\\ &b^{2}-c^{2}=(-6)^{2}-5^{2}=36-25=\color{red}11 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan grafik fungsi kuadrat pada}\\ &\textrm{pada gambar tersebut di atas adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y=6x^{2}-12x+18&&&&&\\ \textrm{B}.&\displaystyle y=6x^{2}+12x+16&&\\ \textrm{C}.&y=6x^{2}-24x+17\\ \textrm{D}.&\color{red}y=6x^{2}-24x+19\\ \textrm{E}.&y=6x^{2}-24x+29 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}+bx+c,\: \: \textrm{dengan}\\ &\textrm{koordinat}\: \: \left ( x_{ss},y_{ss} \right )=(2,-5)\: \: \textrm{dan melalui}\\ &\textrm{titik}\: \: (3,1)\: ,\: \textrm{maka}\\ &y=a(x-x_{ss})^{2}+y_{ss}\Leftrightarrow 1=a(3-2)^{2}+(-5)\\ &\Leftrightarrow 6=a.1^{2}\Leftrightarrow a=6\\ &\textrm{maka persamaan fungsi kuadratnya adalah}:\\ &y=a(x-x_{ss})^{2}+y_{ss}\\ &\Leftrightarrow y=6(x-2)^{2}-5\Leftrightarrow y=6(x^{2}-4x+4)-5\\ &\Leftrightarrow y=\color{red}6x^{2}-24x+19 \end{aligned} \end{array}$

Contoh Soal 1 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 1.&\textrm{Diketahui fungsi}\: \: f(x)=x^{2}-2x-15. \: \: \textrm{Jika}\\ &\textrm{domain}\: \: \left \{ x|-4\leq x\leq 2,x\in \mathbb{R} \right \}\: ,\: \textrm{maka}\\ &range\textrm{-nya adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&-15\leq f(x)\leq 20&&&\\ \textrm{B}.&-15\leq f(x)\leq 9&&\\ \textrm{C}.&\color{red}\displaystyle -16\leq f(x)\leq 9&&\\ \textrm{D}.&\displaystyle -16\leq f(x)\leq 20&&\\ \textrm{E}.&-15\leq f(x)\leq 5&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=x^{2}-2x-15,\: \: \textrm{dengan}\\ &\textrm{D}_{f}=\left \{ x|-4\leq x\leq 2,x\in \mathbb{R} \right \},\\ &\textrm{maka}\: \: range\: \: \textrm{fungsinya adalah}\: \:  \textrm{R}_{f},\: \: \textrm{di mana}\\ &\textrm{R}_{f}\: \: \textrm{diperoleh dengan cara di antaranya}\\ &\textrm{mensubstitusikan langsung ke fungsinya, yaitu}:\\ &f(-4)=(-4)^{2}-2(-4)-15=\color{red}9\\ &f(-3)=(-3)^{2}-2(-3)-15=0\\ &f(-2)=(-2)^{2}-2(-2)-15=-7\\ &f(-1)=(-1)^{2}-2(-1)-15=-12\\ &f(0)=(0)^{2}-2(0)-15=-15\\&f(1)=(1)^{2}-2(1)-15=\color{red}-16\\&f(2)=(2)^{2}-2(2)-15=-15\\ &\textrm{Jadi, range fungsinya}:\textrm{R}_{f}=\color{red}\displaystyle -16\leq f(x)\leq 9 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Daerah hasil fungsi}\: \: f(x)=-x^{2}+6x-5 \: \: \textrm{untuk}\\ &\textrm{daerah asal}\: \: \left \{ x|-1\leq x\leq 6,x\in \mathbb{R} \right \}\: \: \textrm{dan}\\ &y=f(x)\: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\left \{ y|-5\leq y\leq 0,y\in \mathbb{R} \right \}&&&\\ \textrm{B}.&\color{red}\left \{ y|-12\leq y\leq 4,y\in \mathbb{R} \right \}&&\\ \textrm{C}.&\displaystyle \left \{ y|-4\leq y\leq 1,y\in \mathbb{R} \right \}&&\\ \textrm{D}.&\displaystyle \left \{ y|-5\leq y\leq 4,y\in \mathbb{R} \right \}&&\\ \textrm{E}.&\left \{ y|-1\leq y\leq 6,y\in \mathbb{R} \right \}&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Masih sama dengan cara di atas. Diketahui FK}:\\ &f(x)=-x^{2}+6x-5,\: \: \textrm{dengan}\\ &\textrm{D}_{f}=\left \{ x|-1\leq x\leq 6,x\in \mathbb{R} \right \},\\ &\textrm{maka}\: \: range\: \: \textrm{fungsinya adalah}\: \:  \textrm{R}_{f},\: \: \textrm{di mana}\\ &\textrm{R}_{f}\: \: \textrm{diperoleh dengan cara di antaranya}\\ &\textrm{mensubstitusikan langsung ke fungsinya, yaitu}:\\ &f(-1)=-(-1)^{2}+6(-1)-5=\color{red}-12\\ &f(0)=-(0)^{2}+6(0)-5=-5\\ &f(1)=-(1)^{2}+6(1)-5=0\\ &f(2)=-(2)^{2}+6(2)-5=3\\ &f(3)=-(3)^{2}+6(3)-5=\color{red}4\\&f(4)=-(4)^{2}+6(1)-5=3\\&f(5)=-(5)^{2}+6(5)-5=0\\ &f(6)=-(6)^{2}+6(6)-5=-5\\ &\textrm{Jadi, range fungsinya}:\textrm{R}_{f}=\color{red}\left \{ y|-12\leq y\leq 4,y\in \mathbb{R} \right \} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Titik balik parabola}\: \: y=f(x)=-3x^{2}-18x+2\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle (-3,19)&&&&&\textrm{D}.&\displaystyle (3,27)\\ \textrm{B}.&\color{red}\displaystyle (-3,29)&&\textrm{C}.&\displaystyle (-3,23)&&\textrm{E}.&(3,29) \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=-3x^{2}-18x+2\\ &\textrm{Koordinat titik baliknya}=\color{red}\left ( x_{ss},y_{ss} \right )\\ &=\left ( \displaystyle -\frac{b}{2a},-\frac{D}{4a} \right )=\left ( \displaystyle -\frac{b}{2a},-\frac{b^{2}-4ac}{4a} \right )\: \: \textrm{atau}\\ &=\left ( -\displaystyle \frac{b}{2a},f\left ( -\displaystyle \frac{b}{2a} \right ) \right )\\ &=\left ( -\displaystyle \frac{-18}{2(-3)},-\displaystyle \frac{(-18)^{2}-4.(-3).(2)}{4(-3)} \right )\\ &=\color{red}\left (-3,29  \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Fungsi kuadrat dengan titik balik minimum}\\ &(3,-4)\: \: \textrm{dan melalui titik}\: \:  (0,5)\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}y=x^{2}-6x+5&&&\\ \textrm{B}.&y=x^{2}+6x+5&&\\ \textrm{C}.&\displaystyle y=2x^{2}-6x+5&&\\ \textrm{D}.&\displaystyle y=2x^{2}+6x+5&&\\ \textrm{E}.&y=2x^{2}-6x-5&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}\\ &\textrm{Koordinat titik baliknya}=\color{red}\left ( x_{ss},y_{ss} \right )=(3,-4)\\ &\textrm{dan melalui titik}\: \: (0,5),\: \textrm{maka}\\ &5=a(0-\color{red}3\color{black})^{2}+(\color{red}-4\color{black})\Leftrightarrow 5+4=a.9\Leftrightarrow a=\displaystyle \frac{9}{9}=1\\ &\textrm{Sehingga Fk-nya dengan}\: \: a=1\: \: \textrm{adalah}:\\ &f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}=1.(x-3)^{2} +(-4)\\ &\qquad =(x^{2}-6x+9)-4\\ &\qquad=\color{red}x^{2}-6x+5 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Fungsi kuadrat yang melalui titik}\: \: (0,2)\: \: \textrm{dan}\\ &(-1,0)\: \: \textrm{dengan sumbu simetri garis}\\ &x=\displaystyle \frac{1}{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}y=(x+1)(2-x)&&&\\ \textrm{B}.&y=(x-1)(x+2)&&\\ \textrm{C}.&\displaystyle y=2-x-x^{2}&&\\ \textrm{D}.&\displaystyle y=x^{2}-x+2&&\\ \textrm{E}.&y=-(x-1)(x+2)&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}\\ &\textrm{atau}\: \: y=f(x)=a(x-x_{1})(x-x_{2})\: \: \textrm{dengan}\\ &x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{sebagai akar-akarnya}\\ &\textrm{Dan diketahui pula sebagaimana keterangan}\\ &\textrm{dalam soal, maka},\: \: x_{1}=-1,\: x_{ss}=\displaystyle \frac{1}{2}\\ &\textrm{Sehingga}\\ &x_{ss}=-\displaystyle \frac{b}{2a}=\displaystyle \frac{x_{1}+x_{2}}{2}\Leftrightarrow \displaystyle \frac{1}{2}=\frac{-1+x_{2}}{2}\Leftrightarrow x_{2}=2\\ &\textrm{Selanjutnya garfik melalui}\: \: (0,2),\: \textrm{maka}\\ &y=a(x-x_{1})(x-x_{2})\Leftrightarrow 2=a(0-(-1))(0-2)\\ &\Leftrightarrow 2=a(1)(-2)\Leftrightarrow a=-1\\ &\textrm{Sehingga fungsi akan berupa}\\ &f(x)=a(x-x_{1})(x-x_{2})=-1(x+1)(x-2)\\ &\qquad =\color{red}(x+1)(2-x) \end{aligned} \end{array}$.

Contoh Soal 4 Persamaan Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 16.&\textrm{Terdapat dua bilangan bulat positif yang akan}\\ &\textrm{disusun di antara 3 dan 9 sehingga tiga bilangan}\\ &\textrm{pertama membentuk barisan geometri, sedangkan}\\ &\textrm{tiga barisan terakhir membentuk barisan aritmetika}.\\ &\textrm{Jumlah dari dua bilangan tersebut adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&13\displaystyle \frac{1}{2}&&&&&\textrm{D}.&\displaystyle 10\\ \textrm{B}.&\color{red}\displaystyle 11\frac{1}{4}&&\textrm{C}.&\displaystyle 10\frac{1}{2}&&\textrm{E}.&9\displaystyle \frac{1}{2} \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Misalkan bilangan yang dimaksud adalah}:3,x,y,9\\ &\textrm{maka}\\ &\bullet \quad\textrm{Membentuk barisan geometri}:3,x,y\Rightarrow x^{2}=3y\\ &\bullet \quad\textrm{Membentuk barisan aritmetika}:x,y,9\Rightarrow 2y=x+9\\ &\textrm{Selanjutnya}\\ &x^{2}=3y=3\left ( \displaystyle \frac{x+9}{2} \right )\Leftrightarrow 2x^{2}-3x-27=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{3\pm \sqrt{3^{2}+4.2.27}}{2.2}=\displaystyle \frac{3\pm 15}{4}\\ &\textrm{Pilih}\: \: x=\displaystyle \frac{3+15}{4}=\frac{9}{2}\Rightarrow y=\displaystyle \frac{\displaystyle \frac{9}{2}+9}{2}=\frac{27}{4},\\ &\textrm{maka nilai}\: \: x+y=\displaystyle \frac{9}{2}+\frac{27}{4}=\frac{18+27}{4}=\frac{45}{4}=\color{red}11\displaystyle \frac{1}{4}  \end{aligned} \end{array}$

Contoh Soal 3 Persamaan Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{ll}\\ 11.&\textrm{Jumlah kuadrat dari penyelesaian persamaan}\\ &\textrm{kuadrat}\: \: x^{2}+2hx=3\: \: \textrm{adalah 10. Nilai mutlak}\\ &\textrm{dari}\: \: h\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle -1&&&&&\textrm{D}.&\displaystyle 2\\ \textrm{B}.&\displaystyle \displaystyle \frac{1}{2}&&\textrm{C}.&\displaystyle \displaystyle \frac{3}{2}&&\textrm{E}.&\textrm{Salah semua} \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\textrm{Misalkan penyelesaian dari PK}:x^{2}+2hx-3=0\\ &\alpha \: \: \textrm{dan}\: \: \beta ,\: \textrm{maka}\: \: \alpha ^{2}+\beta ^{2}=10\Leftrightarrow (\alpha +\beta )^{2}-2\alpha \beta =10\\ &\Leftrightarrow \left (  -\displaystyle \frac{b}{a}\right )^{2}-2\left ( \displaystyle \frac{c}{a} \right )=10\Leftrightarrow (-2h)^{2}-2(-3)=10\\ &\Leftrightarrow 4h^{2}=10-6=4\Leftrightarrow h^{2}=1\Leftrightarrow \left | h \right |=1\Leftrightarrow h=\pm 1\\ &\textrm{Jadi, nilai yang memenuhi adalah}\: \: \color{red}h=-1 \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: x^{2}+2\left | x \right |-8=0\: ,\: \textrm{maka nilai}\: \: x\\ &\textrm{yang memenuhi adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -4&&&&&\textrm{D}.&\displaystyle 0\\ \textrm{B}.&\color{red}\displaystyle -2&&\textrm{C}.&\displaystyle -1&&\textrm{E}.&4 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &x^{2}+2\left | x \right |-8=0\Leftrightarrow \left ( \left | x \right |+4 \right )\left ( \left | x \right |-2 \right )=0\\ &\Leftrightarrow \left | x \right |=-4\: (\textrm{bukan solusi})\: \: \textrm{atau}\: \: \left | x \right |=2\: (\textrm{solusi})\\ &\textrm{Pilih}\: \: \left | x \right |=2\Rightarrow x=\color{red}\pm 2 \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{akar-akar dari persamaan}\\ &x^{2}-2x=\left | x-1 \right |+5,\: \textrm{maka nilai }\\ &\alpha +\beta\: \:  \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -2&&&&&\textrm{D}.&\displaystyle 1\\ \textrm{B}.&\displaystyle -1&&\textrm{C}.&\displaystyle 0&&\textrm{E}.&\color{red}2 \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &\begin{aligned}&x^{2}-2x=\left | x-1 \right |+5\Leftrightarrow x^{2}-2x-5=\left | x-1 \right |\\ &\color{blue}\textrm{Untuk}\: \: \color{black}x\geq 1,\color{blue}\: \: \textrm{persamaan akan menjadi}\\ &x^{2}-2x-5=x-1\Leftrightarrow x^{2}-2x-x-5+1=0\\ &x^{2}-3x-4=0\Leftrightarrow (x-4)(x+1)=0\\ &\Leftrightarrow x=4\: (\textrm{memenuhi})\: \: \textrm{atau}\: \: x=-1\: (\textrm{tidak})\\ &\color{blue}\textrm{Untuk}\: \: \color{black}x<1,\color{blue}\: \: \textrm{persamaan akan menjadi}\\ &x^{2}-2x-5=1-x\Leftrightarrow x^{2}-2x+x-5-1=0\\ &x^{2}-x-6=0\Leftrightarrow (x-3)(x+2)=0\\ &\Leftrightarrow x=3\: (\textrm{tidak})\: \: \textrm{atau}\: \: x=-2\: (\textrm{memenuhi})\\ &\color{blue}\textrm{Pilih}\: \: \alpha =4,\: \textrm{dan}\: \: \beta =-2,\: \textrm{maka}\\ &\alpha +\beta =4+(-2)=\color{red}2   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Persamaan kuadrat}\: \: x^{2}-2x+m=0 \: \: \textrm{mempunyai }\\ &\textrm{akar-akar yang rasional, maka nilai}\: \: m\: \: \textrm{yang}\\ &\textrm{mungkin adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}1-\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&&\\ \textrm{B}.&1+\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{C}.&\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{D}.&\displaystyle \frac{k^{2}-1}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{E}.&1-\displaystyle \frac{k}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Akar-akar dari PK}:x^{2}-2x+m=0\\ &x_{1,2}=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{2\pm \sqrt{4-4m}}{2}\\  &\textrm{Agar nilai}\: \: m\: \: \textrm{rasional, maka}\\ &4-4m=k^{2}\Leftrightarrow 4m=4-k^{2}\Leftrightarrow m=\color{red}1-\displaystyle \frac{k^{2}}{4}  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Penyelesaian terbesar dikurangi penyelesaian}\\ &\textrm{terkecil dari persamaan kuadrat}\\ &\left ( 7+4\sqrt{3} \right )x^{2}+\left ( 2+\sqrt{3} \right )x-2=0\: \:  \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -2+3\sqrt{3}&&&&&\textrm{D}.&\color{red}\displaystyle 6-3\sqrt{3}\\ \textrm{B}.&\displaystyle 2-\sqrt{3}&&\textrm{C}.&\displaystyle 6+3\sqrt{3}&&\textrm{E}.&3\sqrt{3}+2 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Misalkan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akarnya, maka}\\&\alpha -\beta =\left |\displaystyle \frac{\sqrt{D}}{a}  \right |=\frac{\sqrt{b^{2}-4ac}}{a}\\ &\qquad=\left |\displaystyle \frac{\sqrt{(2+\sqrt{3})^{2}-4(7+4\sqrt{3})(-2)}}{7+4\sqrt{3}}  \right |\\ &\qquad=\left |\displaystyle \frac{\sqrt{4+3+4\sqrt{3}+56+32\sqrt{3}}}{7+4\sqrt{3}}  \right |\\ &\qquad=\left |\displaystyle \frac{\sqrt{63+36\sqrt{3}}}{7+4\sqrt{3}}  \right |=\left |\displaystyle \frac{\sqrt{9(7+4\sqrt{3})}}{7+4\sqrt{3}}   \right |\\ &\qquad=\displaystyle \frac{3}{\sqrt{7+4\sqrt{3}}}=\frac{3}{\sqrt{(2+\sqrt{3})^{2}}}=\displaystyle \frac{3}{2+\sqrt{3}}\\ &\qquad=\displaystyle \frac{3}{2+\sqrt{3}}\frac{2-\sqrt{3}}{2-\sqrt{3}}=3(2-\sqrt{3})=\color{red}6-3\sqrt{3} \end{aligned}  \end{array}$.

Contoh Soal 2 Persamaan Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 6.&\textrm{Diketahui}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{akar-akar dari persamaan}\\ &x^{2}-3x-4=0\: .\: \textrm{Persamaan kuadrat baru yang}\\ &\textrm{memiliki akar-akar}\: \: 2x_{1}\: \: \textrm{dan}\: \: 2x_{2}\: \: \textrm{adalah}\: ....\\  &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 2x^{2}+6x-16=0&&&&&\textrm{D}.&\color{red}\displaystyle x^{2}-6x-16=0\\ \textrm{B}.&\displaystyle 2x^{2}-6x-16=0&&&&&\textrm{E}.&\displaystyle x^{2}+6x-16=0\\ \textrm{C}.&x^{2}+6x+16=0&& \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui bahwa PK}:x^{2}-3x-4=0\\ &\textrm{dengan}\: \: a=1,\: b=-3,\: \: \textrm{dan}\: \: c=-4\\ &\color{blue}\textrm{Alternatif 1}\\ &\textrm{Misalkan Persamaan Kuadrat baru dengan}\\ &\textrm{akar-akar}\: \: \alpha=2x_{1}\: \: \textrm{dan}\: \: \beta =2x_{2},\: \textrm{adalah}\\ &\color{red}x^{2}-(\alpha +\beta )x+\alpha \beta =0\\ &\Leftrightarrow x^{2}-(2x_{1}+2x_{2})x+2x_{1}\times 2x_{2}=0\\ &\Leftrightarrow x^{2}-2(x_{1}+x_{2})x+4x_{1}x_{2}=0\\ &\Leftrightarrow x^{2}-2\left (\displaystyle  -\frac{b}{a} \right )x+4\left ( \displaystyle \frac{c}{a} \right )=0\\ &\Leftrightarrow x^{2}-2(3)x+4(-4)=0\Leftrightarrow \color{red}x^{2}-6x-16=0\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\textrm{PK lama}:x^{2}-3x-4=0\\ &\qquad\qquad\textrm{dengan}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{PK baru dengan}\: \: 2x_{1}\: \: \textrm{dan}\: \: 2x_{2}\\ &\textrm{PK baru}:x^{2}-3(\color{red}2\color{black})x-4(\color{red}2^{2}\color{black})=0\\ &\qquad\qquad\Leftrightarrow  \color{red}x^{2}-6x-16=0\\ &\textrm{Formula tersebut dapat digunakan},\\ &\textrm{syaratnya koefisien dari}\: \: x^{2}=1\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{akar-akar dari persamaan}\\ &2x^{2}-3x+4=0\: .\: \textrm{Persamaan kuadrat baru yang}\\ &\textrm{memiliki akar-akar}\: \: 2x_{1}-1\: \: \textrm{dan}\: \: 2x_{2}-1\: \: \textrm{adalah}\: ....\\  &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 2x^{2}+x-6=0&&&&&\textrm{D}.&\color{red}\displaystyle x^{2}+x-6=0\\ \textrm{B}.&\displaystyle x^{2}+5x+6=0&&&&&\textrm{E}.&\displaystyle x^{2}-x+6=0\\ \textrm{C}.&x^{2}-5x+6=0&& \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui bahwa PK}:2x^{2}-3x+4=0\\ &\textrm{dengan}\: \: a=2,\: b=-3,\: \: \textrm{dan}\: \: c=4\\ &\begin{aligned}&\textrm{Misalkan Persamaan Kuadrat baru dengan}\\ &\textrm{akar-akar}\: \: \alpha=2x_{1}-1\: \: \textrm{dan}\: \: \beta =2x_{2}-1,\: \textrm{adalah}\\ &\color{red}x^{2}-(\alpha +\beta )x+\alpha \beta =0\\ &\Leftrightarrow x^{2}-(2x_{1}-1+2x_{2}-1)x+(2x_{1}-1)\times (2x_{2}-1)=0\\ &\Leftrightarrow x^{2}-\left (2(x_{1}+x_{2})-2  \right )x+4x_{1}x_{2}-2(x_{1}+x_{2})+1=0\\ &\Leftrightarrow x^{2}-\left (2\left (\displaystyle  -\frac{b}{a} \right )-2  \right )x+4\left ( \displaystyle \frac{c}{a} \right )-2\left ( -\displaystyle \frac{b}{a} \right )+1=0\\ &\Leftrightarrow x^{2}-(2(3/2)-2)x+4(4/2)-2(3/2)+1=0\\ &\Leftrightarrow x^{2}-x+(8-3+1)=0\Leftrightarrow \color{red}x^{2}-x+6=0 \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Misalkan Persamaan Kuadrat baru dengan}\\ &\left ( \displaystyle \frac{a+b}{2} \right )\: \textrm{adalah 6 dan rata-rata geometri}\\ & \sqrt{ab}\: \: \textrm{dari kedua bilangan tersebut adalah 10}\\ &\textrm{Persamaan kuadrat yang akar-akarnya kedua}\\ &\textrm{kedua bilangan tersebut adalah}\: ....\\  &\begin{array}{lllllllll}\textrm{A}.&\displaystyle x^{2}+12x-100=0&&&&&\textrm{D}.&\color{red}\displaystyle x^{2}-12x+100=0\\ \textrm{B}.&\displaystyle x^{2}+6x+100=0&&&&&\textrm{E}.&\displaystyle x^{2}-6x+100=0\\ \textrm{C}.&x^{2}-12x-10=0&& \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Formula PK}:x^{2}-(a+b)x+ab=0\\ &\textrm{dengan}\: \: \left\{\begin{matrix} \left ( \displaystyle \frac{a+b}{2} \right )\Rightarrow a+b=12\\ \sqrt{ab}=10\Rightarrow ab=100 \end{matrix}\right.\\ &\textrm{PK yang diinginkan}:\color{red}x^{2}-12x+100=0 \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Akar-akar dari persamaan}\: \: x^{2}+(m-1)x-5=0\\ &\textrm{adalah}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}.\: \textrm{Jika}\: \: x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}=8m,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{ adalah}\: ....\\  &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -6\: \: \textrm{atau}\: \: -14&&&&&\textrm{D}.&\color{red}\displaystyle 3\: \: \textrm{atau}\: \: 7\\ \textrm{B}.&\displaystyle 6\: \: \textrm{atau}\: \: 14&&&&&\textrm{E}.&\displaystyle -3\: \: \textrm{atau}\: \: -7\\ \textrm{C}.&3\: \: \textrm{atau}\: \: -7&& \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Diketahui} \: \: x^{2}+(m-1)x-5=0\\ &\textrm{dengan akar-akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \left\{\begin{matrix} x_{1}+ x_{2}=1-m\\ x_{1}\times  x_{2}=-5\: \: \: \: \: \:  \end{matrix}\right.\\ &\textrm{Selanjutnya},\\ &x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}=8m\\ &\Leftrightarrow \left ( x_{1}+ x_{2} \right )^{2}-4x_{1}x_{2}=8\Leftrightarrow (1-m)^{2}-4(-5)=8m\\ &\Leftrightarrow 1-2m+m^{}+20-8m=0\\ &\Leftrightarrow m^{2}-10m+21=0\Leftrightarrow (m-3)(m-7)=0\\ &\Leftrightarrow m=3\: \: atau\: \: m=7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Agus dan Budi dapat menyelesaikan pengecatan }\\ &\textrm{secara bersama-sama dalam 8 hari. Jika bekerja }\\ &\textrm{sendiri, Budi membutuhkan waktu 12 hari lebih}\\ &\textrm{lama dari Agus. Waktu yang Agus jika ia bekerja}\\ &\textrm{sendiri mengecat rumah tersebut adalah}\: ...\: \textrm{hari}\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 10&&&&&\textrm{D}.&\displaystyle 16\\ \textrm{B}.&\color{red}\displaystyle 12&&\textrm{C}.&\displaystyle 14&&\textrm{E}.&18 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\color{blue}\textrm{Waktu yang dibutuhkan}\\ &\bullet \: \textrm{Waktu yang dibutuhkan Agus}=x\: \: \textrm{hari}\\ &\bullet \: \textrm{Waktu yang dibutuhkan Budi}=x+12\: \: \textrm{hari, dan}\\ &\bullet \: \textrm{Waktu yang dibutuhkan Agus dan Budi}=8\: \: \textrm{hari}\\ &\color{blue}\textrm{Hasil pekerjaan pengecatan rumah dalam sehari}\\ &\bullet \: \textrm{Agus dalam 8 hari}=\displaystyle \frac{8}{x}\: \: \textrm{bagian}\\ &\bullet \: \textrm{Budi dalam 8 hari}=\displaystyle \frac{8}{x+12}\: \: \textrm{bagian, dan}\\ &\bullet \: \textrm{Bagian Agus dan Budi dalam 8 hari}\\&\qquad\qquad\qquad\qquad\qquad \displaystyle \frac{8}{x}+\frac{8}{x+12}=1\\ &\textrm{Sehingga}\\ &\displaystyle \frac{8}{x}+\frac{8}{x+12}=1\Leftrightarrow \frac{8(x+12)+8(x)}{x(x+12)}-1=0\\ &\Leftrightarrow \displaystyle \frac{8x+96+8x-x(x+12)}{x(x+12)}=0\\ &=\Leftrightarrow -x^{2}+4x+96=0\Leftrightarrow x^{2}-4x-96=0\\ &\Leftrightarrow (x-12)(x+8)=0\\ &x=12\: (\textrm{solusi})\: \: \textrm{atau}\: \: x=-8\: (\textrm{bukan})     \end{aligned} \\ &\textrm{Jadi, waktu yang dibutuhkan Agus adalah}\: \: \color{red}\displaystyle 12\: \: \textrm{hari} \end{array}$

Contoh Soal 1 Persamaan Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 1.&\textrm{Penyelesaian terkecil dari persamaan kuadrat}\\ & \left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&-\displaystyle \frac{3}{4}&&&&&\textrm{D}.&\displaystyle \frac{3}{4}\\ \textrm{B}.&\displaystyle \frac{1}{2}&&\textrm{C}.&\color{red}\displaystyle \frac{5}{8}&&\textrm{E}.&1 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4}+x-\frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( 2x-\displaystyle \frac{5}{4} \right )=0\\ &\Leftrightarrow x-\displaystyle \frac{3}{4}=0\: \: \textrm{atau}\: \: 2x-\frac{5}{4}=0\\ &\Leftrightarrow x=\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 2x=\frac{5}{4}\\ &\Leftrightarrow x=\displaystyle \frac{6}{8}\: \: \textrm{atau}\: \: x=\frac{5}{8}\\  \end{aligned} \\ &\textrm{Jadi, nilai terkecilnya adalah}\: \: \color{red}\displaystyle \frac{5}{8} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: m\: \: \textrm{dan}\: \: n\: \: \textrm{adalah penyelesaian dari}\\ &\textrm{persamaan kuadrat}\: \: x^{2}+mx+n=0,\\ &\textrm{dengan}\: \: m\neq 0\: \: \textrm{dan}\: \: n\neq 0\: \: \textrm{jumlah kedua}\\ &\textrm{penyelesaian tersebut adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -\frac{1}{2}&&&&&\textrm{D}.&\displaystyle 1\\ \textrm{B}.&\color{red}\displaystyle -1&&\textrm{C}.&\displaystyle \frac{1}{2}&&\textrm{E}.&\displaystyle \textrm{tidak dapat ditentukan} \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa PK}:x^{2}+mx+n=0\\ &\textrm{dengan}\: \: a=1,\: b=m,\: \: \textrm{dan}\: \: c=n\\ &\begin{aligned}&\bullet \quad x_{1}+ x_{2}=-\displaystyle \frac{b}{a}\Leftrightarrow m+n=-\frac{m}{1}\\ &\bullet \quad x_{1}\times x_{2}=\displaystyle \frac{c}{a}\quad\Leftrightarrow mn=\frac{n}{1}\Leftrightarrow m=1\\ &\textbf{Dari persamaan pertama akan diperoleh}\\ &m+n=-m=\color{red}-1 \end{aligned}\\ &\textrm{Jadi, nilai m+n adalah}\: \: \color{red}\displaystyle -1 \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika persamaan}\: \: 2x^{2}-3x-14=0\: \: \textrm{mempunyai}\\&\textrm{akar-akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{dengan}\: \: x_{1}>x_{2}, \: \: \textrm{maka}\\ &2x_{1}+3x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 2&&&&&\textrm{D}.&\displaystyle -2\\ \textrm{B}.&\color{red}\displaystyle 1&&\textrm{C}.&\displaystyle -1&&\textrm{E}.&-5 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa PK}\: :\: 2x^{2}-3x-14=0\\ &\begin{aligned}&2x^{2}-3x-14=0\Leftrightarrow \displaystyle \frac{(2x-7)(2x+4)}{2}=0\\ &\Leftrightarrow (2x-7)(x+2)=0\\ &\Leftrightarrow 2x-7=0\: \: \textrm{atau}\: \:  x+2=0\\ &x=\displaystyle \frac{7}{2}\: \: \textrm{atau}\: \:  x=-2\\ &\textrm{Karena nilai dari}\: \: x_{1}>x_{2},\: \: \textrm{maka}\\ &x_{1}=\displaystyle \frac{7}{2}\: \: \textrm{dan}\: \:  x_{2}=-2\\ \end{aligned}\\ &\textrm{Sehingga nilai}\: \: 2x_{1}+3x_{2}=\color{blue}2\displaystyle \frac{7}{2}+3(-2)\color{black}=7-6=\color{red}1 \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{adalah akar-akar dari}\\ &\textrm{persamaan kuadrat}\: \: x^{2}+6x+2=0,\\ &\textrm{nilai dari}\: \: x_{1}^{2}+x_{2}^{2}-4x_{1}x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 28&&&&&\textrm{D}.&\displaystyle 18\\ \textrm{B}.&\displaystyle 26&&\textrm{C}.&\color{red}\displaystyle 24&&\textrm{E}.&\displaystyle 16 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\textrm{Diketahui bahwa PK}:x^{2}+6x+2=0\\ &\textrm{dengan}\: \: a=1,\: b=6,\: \: \textrm{dan}\: \: c=2\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\color{blue}x_{1}^{2}+x_{2}^{2}\color{black}-4x_{1}x_{2}\\ &=\color{blue}\left ( x_{1}+x_{2} \right )^{2}-2x_{1}x_{2}\color{black}-4x_{1}x_{2}\\ &=\left ( x_{1}+x_{2} \right )^{2}-6x_{1}x_{2}\\ &=\left ( -\displaystyle \frac{b}{a} \right )^{2}-6\left ( \displaystyle \frac{c}{a} \right )=(-6)^{2}-6(2)\\ &=36-12\\ &=\color{red}24 \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&x^{2}+6x+2=0\Leftrightarrow x^{2}=-6x-2\\ &\bullet \quad x=x_{1}\Rightarrow x_{1}^{2}=-6x_{1}-2\\ &\bullet \quad x=x_{2}\Rightarrow x_{2}^{2}=-6x_{2}-2\\ &\qquad \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\quad+ \\ & \Leftrightarrow \quad x_{1}^{2}+x_{2}^{2}=-6\left (x_{1}+x_{2}  \right )-4\\ & \Leftrightarrow \quad x_{1}^{2}+x_{2}^{2}-4x_{1}x_{2}=-6\left (x_{1}+x_{2}  \right )-4x_{1}x_{2}-4\\  &\qquad\qquad\qquad=-6\left ( -\displaystyle \frac{b}{a} \right )-4\left ( \frac{c}{a} \right )-4\\ &\qquad\qquad\qquad=-6(-6)-4(2)-4\\ &\qquad\qquad\qquad=36-8-4\\ &\qquad\qquad\qquad=\color{red}24\\ \end{aligned}\\ &\textrm{Jadi, nilai yang dimaksud adalah}\: \: \color{red}\displaystyle 24 \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diketahui akar-akar dari persamaan}\: \: 7x=4x^{2}+3\\ &\textrm{adalah}\: \: \alpha \: \: \textrm{dan}\: \: \beta .\: \textrm{Nilai} \: \: \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=....\\  &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \frac{25}{12}&&&&&\textrm{D}.&\displaystyle \frac{16}{25}\\\\ \textrm{B}.&\displaystyle \frac{24}{12}&&\textrm{C}.&\displaystyle \frac{20}{25}&&\textrm{E}.&\displaystyle \frac{12}{25} \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\textrm{Diketahui bahwa PK}:4x^{2}-7x+3=0\\ &\textrm{dengan}\: \: a=4,\: b=-7,\: \: \textrm{dan}\: \: c=3\\ &\begin{aligned}&\displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\displaystyle \frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }=\frac{(\alpha +\beta )^{2}-2\alpha \beta }{\alpha \beta }\\ &=\displaystyle \frac{\left (\displaystyle -\frac{b}{a} \right )^{2} -2\left (\displaystyle \frac{c}{a}  \right )}{\displaystyle \frac{c}{a}}=\displaystyle \frac{\left ( \displaystyle \frac{7}{4} \right )^{2}-2\left ( \displaystyle \frac{3}{4} \right )}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{\displaystyle \frac{49}{16}-\frac{6}{4}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{\displaystyle \frac{49-24}{16}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{\displaystyle \frac{25}{16}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{25}{16}\times \frac{4}{3}\\ &=\color{red}\displaystyle \frac{25}{12} \end{aligned} \end{array}$.

Sumber Referensi:

1. Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.

Fungsi Kuadrat (Kelas X/Fase E Semester 2)

 B. Fungsi Kuadrat

B. 1 Fungsi

Silahkan lihat materi sebelumnya, cari di blog ini

di sini 1, dan di sini 2

B. 2 Fungsi Kuadrat

Perhatikan tabel berikut

$\begin{array}{|l|l|}\hline \textrm{Pengertian}&\begin{aligned}&\textrm{Suatu fungsi yang berbentuk}\\ &f(x)=ax^{2}+bx+c\\ & a,\: b,\: c,\: \in \mathbb{R},\: a\neq 0 \end{aligned}\\\hline \textrm{Grafik Fungsi}&\textrm{Keterangan}\\\hline \textrm{Titik potong sumbu x}&\textrm{Jika ada}\\\hline &\begin{aligned}&\textrm{untuk titik potong}\\ &\textrm{terhadap sumbu x }\\ &\textrm{Jika y = 0 maka }\\ &ax^{2}+bx+c=0\\ &\textrm{Selanjutnya tinggal}\\ &\textrm{menentukan nilai D}\\ &D=b^{2}-4ac\: \: \textrm{adalah}\\ &\: \: \: \: \: \: \: \: \: \textrm{nilai diskriminan}.\\ &\textrm{Jika} \: D>0\\ &\textrm{maka grafik}\\ &\textrm{memotong sumbu x}\\ &\textrm{di dua tempat berbeda}\\ &\textrm{yaitu di} \: (x_{1},0)\: \textrm{dan}\: (x_{2},0).\\ &\textrm{dan jika D = 0}\\ &\textrm{maka grafik}\\ &\textrm{ hanya menyinggung}\\ &\textrm{sumbu x di satu titik}\\ &\textrm{yaitu di }\: (x_{1},0)\\ &\textrm{dan jika}\: D<0 \\ &\textrm{maka grafik}\\ &\textrm{tidak memotong}\\ &\textrm{atau menyinggung sumbu x} \end{aligned}\\\hline \textrm{Titik potong sumbu y}&\begin{aligned}&\textrm{titik potong terhadap}\\ &\textrm{sumbu y, jika x = 0}\\ &y=f(x)=ax^{2}+bx+c\\ &y=f(0)=a(0)^{2}+b(0)+c\\ &y=c \end{aligned}\\\hline \textrm{Sumbu Simetri (SS)}&x=\displaystyle \frac{-b}{2a}\\\hline \textrm{Titik Puncak}&\left ( \displaystyle \frac{-b}{2a},\displaystyle \frac{D}{-4a} \right )\\\hline \textrm{Posisi grafik}&\textrm{Jika}\: a>0\: \textrm{maka}\\ &\textrm{grafik terbuka ke atas}\\ &\textrm{Dan jika nilai}\: a<0\: \textrm{maka}\\ &\textrm{grafik terbuka ke bawah}\\\hline \end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|c|c|}\hline \textrm{Jika memotong sumbu}-\textrm{X}&\textrm{Jika menyinggung sumbu}-\textrm{X}\\ \textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan}\: \left ( x_{2},0 \right )&\textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan melalui}\\ \textrm{dan melalui sebuah titik lain}&\textrm{sebuah titik lain} \\\hline &\\ y=f(x)=a\left ( x-x_{1} \right )\left ( x-x_{2} \right )&y=f(x)=a\left ( x-x_{1} \right )^{2}\\ &\\\hline \textrm{Jika grafik fungsi itu melalui}&\textrm{Jika grafik fungsi itu melalui}\\\hline \textrm{Titik puncak}\: \: P\left ( x_{p},y_{p} \right )\: \textrm{dan}&\textrm{tiga buah titik yaitu}\: \left ( x_{1},y_{1} \right )\\ \textrm{sebuah titik lain}&\left ( x_{2},y_{2} \right )\: \: \textrm{dan}\: \: \left ( x_{3},y_{3} \right )\\\hline &\\ y=f(x)=a\left ( x-x_{p} \right )^{2}+y_{p}&y=f(x)=ax^{2}+bx+c\\ &\\\hline \end{array}$.

B. 3 Masalah yang Melibatkan Fungsi Kuadrat

$\begin{aligned}&y=f(x)=ax^{2}+bx+c\\ &\quad \textrm{dengan}\: a,b,c\in \mathbb{R},\: a\neq 0 \end{aligned}$.

B.3.1 Titik Stasioner

$\begin{aligned}&y_{ekstrim}=\color{red}-\displaystyle \frac{D}{4a}\color{black}\Rightarrow \left\{\begin{matrix} y_{\textrm{minimum}},\: \textrm{jika}\: a>0\\  y_{\textrm{maksimum}},\: \textrm{jika}\: a<0 \end{matrix}\right.\\ &y_{\textrm{ekstrim}}\: \textrm{tercapai saat}\: \: x=\color{red}-\displaystyle \frac{b}{2a}\\ &\textbf{Sehingga titik stasionernya adalah}\\ &\qquad\qquad\qquad =\left ( x_{ss},y_{ss} \right )=\left ( -\displaystyle \frac{b}{2a},-\displaystyle \frac{D}{4a} \right ) \end{aligned}$.

B.3.2 Definit Positif dan Definit Negatif

$\begin{aligned}&\textrm{Jika}\: \: D<0\: \: \textrm{dan}\left\{\begin{matrix} a>0,\: \: \textrm{maka}\: \: y\: \: \textrm{akan selalu positif}\\  a<0,\: \: \textrm{maka}\: \: y\: \: \textrm{akan selalu negatif} \end{matrix}\right.\\ &\textrm{untuk setiap nilai}\: \: x \end{aligned}$.

Perhatikan tambahan penjelasan berikut

$\begin{aligned}&\textrm{Tentang definit positif dan negatif}\\ &\begin{array}{cccc}\\a>0.D<0&\textrm{Gambar}&\LARGE\cup &\textbf{Sumbu-X}\\\hline a<0,D<0&\textrm{Gambar}&\cap &\\ & \end{array} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: f\: \: \textrm{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \textrm{maka nilai dari}\: \: f(4)-f(-2)\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &f(x)=ax+b\\ &f(2)-f(-2)\\ &=\left (a(2)+b \right )-\left ( a(-2)+b \right )=8\\ &8=2a+2a\\ &8=4a\\ &2=a\\ &f(x)=2x+b,\quad \textrm{dengan}\: \: b\: \: \textrm{konstan}\\ &\textrm{Sehingga nilai}\quad\\ &f(4)-f(-2)=\left (2(4)+b \right )-\left (2(-2)+b \right )\\ &=8+b+4-b\\ &=\color{red}12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah}\: \: 8-6x-x^{2}\: \: \textrm{ke dalam bentuk}\\ & a-(x+b)^{2},\: \textrm{selanjutnya tentukan}\\ & \textrm{daerah hasil dari}\: \: f(x)=8-6x-x^{2}\\ & \textrm{untuk}\: \: x\: \: \textrm{bilangan real}\\ &\qquad(\textit{NTU Entrance Examination AO-level})\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|l|}\hline 1.&\color{blue}\textrm{Diketahui}\\\hline &\begin{aligned}\textrm{Misal}\quad\qquad&\\ 8-6x-x^{2}&=f(x)\\ f(x)&=-x^{2}-6x+8\\ &=-\left ( x^{2}+6x-8 \right )\\ &=-\left ( x^{2}+6x+9-17 \right )\\ &=-\left ( (x+3)^{2}-17 \right )\\ &=-(x+3)^{2}+17\\ & \end{aligned}\\\hline 2.&\color{blue}\textrm{Mencari koordinat}\: \: \left ( x_{SS},y_{SS} \right )\\\hline &\begin{aligned}f(x)&=-x^{2}-6x+8\left\{\begin{matrix} a=-1\\ b=-6\\ c=\: \: 8\: \: \end{matrix}\right.\\ \textrm{Maka}&\\ x_{SS}&=\frac{-b}{2a}=\displaystyle \frac{-(-6)}{2(-1)}\\ &=-3\\ y_{SS}&=f(-3)=-\left ( -3+3 \right )^{2}+17=17\\ \therefore &\left ( x_{SS},y_{SS} \right )=(-3,17) \end{aligned}\\\hline 3.&\color{blue}\textrm{Nilai fungsi}\\\hline &\begin{aligned}\textrm{Karena}&\: \: a=-1<0\\ \textrm{maka f}&\textrm{ungsi menghadap}\\ \textbf{ke ba}&\textbf{wah},\: \: \textrm{sehingga}\\ \textrm{daerah}&\: \: \textrm{hasilnya}\: \: \left (R_{f} \right )\\ \textrm{adalah}&:\\ &\left \{ -\infty <y\leq 17 \right \}\\ &\\ &\textrm{Berikut ilustrasinya} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari }\\ &\textrm{persamaan kuadrat}\: \: x^{2}+mx+m=0,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{yang menyebabkan }\\ &\textrm{jumlah kuadrat akar-akar mencapai}\\ &\textrm{minimum adalah}\: ....\\ &\qquad \: \textbf{(UM UNDIP 2014 Mat Das)}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \: x^{2}+mx+m=0\\ & \textbf{persamaan kuadrat}\: \textrm{dalam}\: \: x,\\ & \textrm{maka}\\ &x^{2}+mx+m=x^{2}-(\alpha +\beta )x+(\alpha \beta )=0\\ &\begin{cases} \alpha +\beta &=-m \\ & \\ \alpha \beta &=m \end{cases}\\ &\textrm{Selanjutnya}\\ &\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta\\ &=(-m)^{2}-2m\: \: \textrm{dan dapat kita tuliskan sebagai}\\ &f(m)=m^{2}-2m\begin{cases} a &=1 \\ b &=-2 \\ c &=0 \end{cases} \\ &\textrm{fungsi kuadrat dalam}\: \: m,\\ &\textrm{sehingga kita perlu mencari titik}\: \: \left ( m_{SS},f\left ( m_{SS} \right ) \right ),\\ & \textrm{tetapi yang kita perlukan}\\ &\textrm{cuma}\: \: m-\textrm{nya saja, yaitu}:\: \: m=m_{SS},\\ &\textrm{dengan}\quad m_{SS}=\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1 \end{aligned} \end{array}$.

Sumber Referensi:

1. Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.
2. Kurnianingsih, Sri, Kuntarti & Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1. Jakarta: ESIS.
3. Noormandiri. 2022. Matematika untuk SMA/MA Kelas X.Jakarta: ERLANGGA
4. Sobirin. 2005. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 1. Jakarta: KAWAN PUSTAKA.

Persamaan Kuadrat (Kelas X/Fase E Semester 2)

  Semester Genap

• Persamaan dan Fungsi Kuadrat
• Statistika
• Aturan Pencacahan dan Peluang

A. Persamaan dan Fungsi Kuadrat

A. 1  Bentuk umum persamaan kuadrat

$\begin{aligned}&\color{red}\mathbf{ax^{2}+bx+c=0}\\ &\textrm{dengan}\: \: a,b,c\in \mathbb{R},\: \color{blue}a\neq 0 \end{aligned}$.

Adapun cara penyelesaian persamaan kuadrat, jika $x_{1}\: \: dan\: \: x_{2}$ sebagai akar-akarya adalah:

$\begin{array}{|l|l|l|}\hline \qquad \textbf{Pemfaktoran}&\textrm{Melengkapkan}&\textbf{Rumus ABC}\\ &\textbf{kuadrat sempurna}&\\\hline \qquad\qquad(1)&\qquad\qquad(2)&\qquad\qquad(3)\\\hline \begin{aligned}&ax^{2}+bx+c=0\\ &\left ( x-x_{1} \right )\left ( x-x_{2} \right )=0\\ &\textrm{Jika koefisien}\: \: x^{2}\\ &\textrm{lebih dari 1, maka}\\ &\color{blue}\textrm{ubahlah ke bentuk}\\ &\displaystyle \frac{1}{a}\left ( ax-x_{1} \right )\left ( ax-x_{2} \right )\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&ax^{2}+bx+c=0\\ &x^{2}+\displaystyle \frac{b}{a}x+\frac{c}{a}=0\\ &x^{2}+\displaystyle \frac{b}{a}x=-\frac{c}{a}\\ &\textbf{selanjutnya}\\ &x^{2}+\displaystyle \frac{b}{a}x+\left (\frac{b}{2a}  \right )^{2}\\ &\quad =-\displaystyle \frac{c}{a}+\left (\frac{b}{2a}  \right )^{2}\\ &\left ( x+\displaystyle \frac{b}{2a} \right )^{2}\\ &\quad =\displaystyle \frac{b^{2}-4ac}{4a^{2}} \end{aligned}&\begin{aligned}&\textrm{Dari bentuk 2, kita}\\ &\textrm{akan mendapatkan}\\ &\left ( x+\displaystyle \frac{b}{2a} \right )^{2}\\ &\quad =\displaystyle \frac{b^{2}-4ac}{4a^{2}}\\ &x-\displaystyle \frac{b}{2a}=\pm \sqrt{\displaystyle \frac{b^{2}-4ac}{4a^{2}}}\\ &x=-\displaystyle \frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{\sqrt{4a^{2}}}\\ &x=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ &\\ &\\ \end{aligned} \\\hline \end{array}$.

A. 2.  Jenis-Jenis Akar Persamaan Kuadrat

Pada kondisi ini, akar-akar dari persamaan kuadrat tergantung pada nilai di bawah tanda akar yang selanjutnya dikenal dengan nilai Diskriminan yang selanjutnya disingkat dengan huruf D, dengan nilai $D=b^{2}-4ac$.

$\begin{array}{|c|c|l|}\hline \textrm{No}&\textrm{Jenis nilai}\: \: \textbf{D}&\textrm{Penjelasan nilai}\: \: \textbf{D}\\\hline  1&\textbf{D}>0&\textrm{Persamaan kuadrat mempunyai dua akar }\\ &&\textrm{riil dan}\: \: \color{red}\textrm{berbeda}\\\hline 2&\textbf{D}=0&\textrm{Persamaan kuadrat mempunyai dua akar }\\ &&\textrm{riil dan}\: \: \textit{sama}\\\hline 3&\textbf{D}<0&\textrm{Persamaan kuadrat mempunyai dua akar }\\ &&\textrm{tidak riil dan}\: \: \color{purple}\textrm{berbeda}\\\hline \end{array}$.

A. 3  Jumlah dan Hasil Kali serta Selisih Akar-Akar Persamaan Kuadrat

$\begin{array}{|c|c|l|}\hline \textrm{No}&\textrm{Kondisi akar-akar}\: \: x_{1}\:  \&\: x_{2}&\textrm{Dari posisi}\: \: \color{red}ax^{2}+bx+c=0\\\hline  1&x_{1}+x_{2}=-\displaystyle \frac{b}{a}&\textrm{akar-akarnya tidak harus}\: \: \: x_{1}\:  \&\: x_{2}\\ &&\textrm{terkadang dituliskan dengan}\: \: \color{red}\alpha \: \: \color{black}\textrm{dan}\: \color{red}\: \beta \\\hline 2&x_{1}\times x_{2}=\displaystyle \frac{c}{a}&\textrm{Baik rumus jumlah maupun hasil kali}\\ &&\textrm{Anda juga dapat melihat dari jenis akarnya}\\\hline 3&x_{1}-x_{2}=\left | \displaystyle \frac{\sqrt{D}}{a} \right |&\textrm{Ingat nilai}\: \: D=\color{purple}b^{2}-4ac\\\hline \end{array}$.

A. 4.  Menyusun Persamaan Kuadrat Baru

Persamaan kuadrat dengan akar-akar $x_{1}\: \: dan\: \: x_{2}$  dapat disusun dengan rumus:

$\LARGE x^{2}-\left ( x_{1}+x_{2} \right )x+x_{1}\times x_{2}=0$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan kar-akar dari persamaan kuadrat}\\ &(\textrm{a})\quad x^{2}-2x-8=0\\ &(\textrm{b})\quad 2x^{2}-3x-5=0\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline (\textbf{a})&(\textbf{b})\\\hline \begin{aligned}&x^{2}\color{red}-2\color{black}x-8=0\\ &\Leftrightarrow (x\color{red}-4\color{black})(x\color{red}+2\color{black})=0\\ &\Leftrightarrow x-4=0\: \: \textrm{atau}\: \: x+2=0\\ &\Leftrightarrow x=4\: \: \textrm{atau}\: \: x=-2\\ &\\ &\\ & \end{aligned}&\begin{aligned}&2x^{2}\color{red}-3\color{black}x-5=0\\ &\Leftrightarrow \displaystyle \frac{ (2x\color{red}-5\color{black})(2x\color{red}+2\color{black})}{\color{blue}2}=0\\ &\Leftrightarrow (2x-5)(\color{blue}x+1\color{black})=0\\ &\Leftrightarrow 2x-5=0\: \: \textrm{atau}\: \: x+1=0\\ &\Leftrightarrow 2x=5\: \: \textrm{atau}\: \: x=-1\\ &\Leftrightarrow x=\displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x=-1 \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Penyelesaian terkecil dari persamaan kuadrat}\\ & \left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\left ( \color{red}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( \color{red}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{red}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4}+x-\frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{red}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( 2x-\displaystyle \frac{5}{4} \right )=0\\ &\Leftrightarrow x-\displaystyle \frac{3}{4}=0\: \: \textrm{atau}\: \: 2x-\frac{5}{4}=0\\ &\Leftrightarrow x=\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 2x=\frac{5}{4}\\ &\Leftrightarrow x=\displaystyle \frac{6}{8}\: \: \textrm{atau}\: \: x=\frac{5}{8}\\  \end{aligned}\\ &\textrm{Jadi, nilai terkecilnya adalah}\: \: \color{blue}\displaystyle \frac{5}{8} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika persamaan}\: \: 2x^{2}-3x-14=0\: \: \textrm{mempunyai}\\&\textrm{akar-akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{dengan}\: \: x_{1}>x_{2}, \: \: \textrm{maka}\\ &2x_{1}+3x_{2}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa PK}\: :\: 2x^{2}-3x-14=0\\ &\begin{aligned}&2x^{2}-3x-14=0\Leftrightarrow \displaystyle \frac{(2x-7)(2x+4)}{2}=0\\ &\Leftrightarrow (2x-7)(x+2)=0\\ &\Leftrightarrow 2x-7=0\: \: \textrm{atau}\: \:  x+2=0\\ &x=\displaystyle \frac{7}{2}\: \: \textrm{atau}\: \:  x=-2\\ &\textrm{Karena nilai dari}\: \: x_{1}>x_{2},\: \: \textrm{maka}\\ &x_{1}=\displaystyle \frac{7}{2}\: \: \textrm{dan}\: \:  x_{2}=-2\\ \end{aligned}\\ &\textrm{Sehingga nilai}\: \: 2x_{1}+3x_{2}=\color{red}2\displaystyle \frac{7}{2}+3(-2)\color{black}=7-6=\color{blue}1 \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahu}i\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari }\\ &\textrm{persamaan kuadrat}\: \: x^{2}-x-2=0,\\ &\textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \alpha +\beta \: \: \textrm{dan}\: \: \alpha \beta&\textrm{e}.\quad \alpha ^{2}+\beta ^{2}\\ \textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&\textrm{f}.\quad \alpha ^{2}-\beta ^{2}\\ \textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}\\ \textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}\\ \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}-x-2=0\left\{\begin{matrix} \alpha \\ \\ \beta \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=-1\\ c=-2 \end{matrix}\right.\\ & \end{aligned}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\alpha +\beta =-\frac{b}{a}=-\frac{(-1)}{1}=1\\ &\alpha \beta =\frac{c}{a}=\frac{(-2)}{1}=-2 \end{aligned}&\begin{aligned}\textrm{e}.\quad \alpha ^{2}+\beta ^{2}&=\left ( \alpha +\beta \right )^{2}-2\alpha \beta \\ &=1^{2}-2(-2)\\ &=1+4=5 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&=\frac{D}{a^{2}}\\ &=\frac{b^{2}-4ac}{a^{2}}\\ &=\frac{(-1)^{2}-4.(1).(-2)}{(1)^{2}}\\ &=1+8=9 \end{aligned}&\begin{aligned}\textrm{f}.\quad \alpha ^{2}-\beta ^{2}&=\left ( \alpha +\beta \right )\left ( \alpha -\beta \right )\\ &=(1).(9)=9\\ &\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&=\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }\\ &=\displaystyle \frac{5}{-2}\\ &=-\frac{5}{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}&=\frac{(\alpha -2)+(\beta -2)}{(\alpha -2).(\beta -2)}\\ &=\displaystyle \frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4}\\ &=\displaystyle \frac{(-1)-4}{(-2)-2(-1)+4}\\ &=\displaystyle \frac{-5}{-2+2+4}\\ &=-\frac{5}{4} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&=\frac{\alpha +\beta }{\alpha \beta }\\ &=\displaystyle \frac{(-1)}{(-2)}\\ &=\frac{1}{2} \end{aligned}&\begin{aligned}\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}&=\frac{\alpha ^{3}+\beta ^{3}}{(\alpha \beta )^{2}}\\ &=\displaystyle \frac{(\alpha +\beta )^{3}-3\alpha \beta (\alpha +\beta )}{(\alpha \beta )^{2}}\\ &=.... \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diketahu}i\: \: p \: \: \textrm{dan}\: \: q \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\\ &x^{2}+2x-5=0,\: \: \textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad p^{2}+q^{2}&\textrm{e}.\quad (p-3)^{2}+(q-3)^{2}\\ \textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&\textrm{f}.\quad p^{2}q+pq^{2}\\ \textrm{c}.\quad p^{3}+q^{3}&\textrm{g}.\quad (p+q)^{2}-(p-q)^{2}\\ \textrm{d}.\quad p^{3}-q^{3}&\textrm{h}.\quad (p^{3}+q^{3})-(p^{3}-q^{3}) \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}+2x-5=0\left\{\begin{matrix} p \\ \\ q \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=2\\ c=-5 \end{matrix}\right.\\ & \end{aligned}\\ &\begin{array}{|l|l|}\hline  \begin{aligned}\textrm{a}.\quad p^{2}+q^{2}&=(p+q)^{2}-2pq\\ &=(-\frac{b}{a})^{2}-2\left ( \frac{c}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{2}-2\left ( \frac{(-5)}{1} \right )\\ &=4+10\\ &=14 \end{aligned}&\begin{aligned}\textrm{e}.\quad &(p-3)^{2}+(q-3)^{2}\\ &=p^{2}-6p+9+q^{2}-6q+9\\ &=p^{2}+q^{2}-6(p+q)+18\\ &=14-6(-2)+18\\ &=14+12+18\\ &=44 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&=\frac{p^{2}+q^{2}}{pq}\\ &=\displaystyle \frac{14}{-5}\\ &=-\frac{14}{5} \end{aligned}&\begin{aligned}\textrm{f}.\quad p^{2}q+pq^{2}&=pq(p+q)\\ &=(-5)(-2)\\ &=10\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad p^{3}+q^{3}&=(p+q)^{3}-3pq(p+q)\\ &=\left ( -\frac{b}{a} \right )^{3}-3\left ( \frac{c}{a} \right )\left ( -\frac{b}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{3}-3\left ( \frac{(-5)}{1} \right )\left ( -\frac{2}{1} \right )\\ &=-8-30\\ &=-38\\ & \end{aligned}&\begin{aligned}\textrm{d}.\quad p^{3}&-q^{3}\\ &=(p-q)^{3}+3pq(p-q)\\ &=\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )^{^{3}}+3\left ( \frac{c}{a} \right )\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )\\ &=\left ( \displaystyle \frac{\sqrt{2^{2}-4.1.(-5)}}{1} \right )^{3}+3\left ( \frac{-5}{1} \right )....\\ &=....\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukanlah akar-akar persamaan kuadrat dengan rumus kuadrat}\\ &\begin{array}{ll}\\ \textrm{a}.\quad x^{2}-2=0&\textrm{f}.\quad 2p^{2}-5p-12=0\\ \textrm{b}.\quad x^{2}+3x-1=0&\textrm{g}.\quad 3q^{2}-11q+10=0\\ \textrm{c}.\quad x^{2}+2x-3=0&\textrm{h}.\quad 4x^{2}+11x+6=0\\ \textrm{d}.\quad x^{2}+5x-6=0&\textrm{i}.\quad 5z^{2}-z-4=0\\ \textrm{e}.\quad x^{2}-7x-8=0&\textrm{j}.\quad 6x^{2}+17x+7=0 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad x^{2}&-2=0\\ &\begin{cases} a & =1 \\ b & =0 \\ c & =-2 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x_{1,2}&=\displaystyle \frac{0\pm \sqrt{0^{2}-4.1.(-2)}}{2.1}\\ &=\displaystyle \frac{\pm \sqrt{8}}{2}=\frac{\pm \sqrt{4.2}}{2}\\ &=\displaystyle \frac{\pm 2\sqrt{2}}{2}\\ &=\pm \sqrt{2}\\ x_{1}&=\sqrt{2}\quad \textrm{atau}\quad x_{2}=-\sqrt{2}\end{aligned}&\begin{aligned}\textrm{i}.\quad 5z^{2}&-z-4=0\\ &\begin{cases} a & =5 \\ b & =-1 \\ c & =-4 \end{cases}\\ z_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ z_{1,2}&=\displaystyle \frac{-(-1)\pm \sqrt{(-1)^{2}-4.5.(-4)}}{2.5}\\ &=\displaystyle \frac{1\pm \sqrt{1+80}}{10}=\frac{1\pm \sqrt{81}}{10}\\ &=\displaystyle \frac{1\pm 9}{10}\\ z_{1}&=\displaystyle \frac{1+9}{10}=1\quad \textrm{atau}\quad z_{2}=\frac{1-9}{10}=\frac{-8}{10}=-\frac{4}{5}\\ &\end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa untuk}\: \: m\in \textrm{rasional},\: \textrm{maka kedua akar persamaan}\\ &\textrm{a}.\quad x^{2}+(m+2)x+2m=0,\: \textrm{adalah rasional juga}\\ &\textrm{b}.\quad 2x^{2}+(m+4)x+(m-1)=0,\: \textrm{selalu memiliki dua akar real yang berlainan}\\ &\textrm{c}.\quad x^{2}+(m+4)x-2m^{2}-m+3=0,\: \textrm{selalu memiliki dua akar real dan rasional}\\\\ &\textbf{Bukti}:\\ &\begin{array}{|c|c|c|}\hline  x^{2}+(m+2)x+2m=0&2x^{2}+(m+4)x+(m-1)=0&x^{2}+(m+4)x-2m^{2}-m+3=0\\\hline \begin{aligned}a&=1,\: b=(m+2),\: c=2m \end{aligned}&\begin{aligned}a&=2,\: b=m+4,\: c=m-1 \end{aligned}&\begin{aligned}a&=1,\: b=m+4,\: c=-2m^{2}-m+3 \end{aligned}\\\hline \begin{aligned}D&=(m+2)^{2}-4.1.(2m)\\ &=m^{2}+4m+4-8m\\ &=m^{2}-4m+4\\ &=(m-2)^{2} \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.2.(m-1)\\ &=m^{2}+8m+16-8m+8\\ &=m^{2}+24\\ & \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.1.(-2m^{2}-m+3)\\ &=m^{2}+8m+16+8m^{2}+4m-12\\ &=9m^{2}+12m+4\\ &=(3m+2)^{2} \end{aligned}\\\hline  \begin{aligned}&\textrm{2 akar rasional} \end{aligned}&\begin{aligned}&\textrm{2 akar real dan berbeda} \end{aligned}&\begin{aligned}&\textrm{2 akar rasional} \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Carilah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}&=\frac{2}{x^{2}-10x-69}\\ \displaystyle \frac{1}{\left (x^{2}-10x-37 \right )+8}+\frac{1}{\left (x^{2}-10x-37 \right )-8}&=\frac{2}{\left (x^{2}-10x-37 \right )-32}\\ \textrm{Misalkan}\: \: x^{2}-10x-37=p, \: \: \textrm{maka}\qquad&\\ \displaystyle \frac{1}{p+8}+\frac{1}{p-8}&=\frac{2}{p-32}\\ \displaystyle \frac{p-8+p+8}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{2p}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{p}{p^{2}-64}&=\frac{1}{p-32}\\ p^{2}-32p&=p^{2}-64\\ p&=\displaystyle \frac{-64}{-32}\\ p&=2,\\ \textnormal{kita kembali ke bentuk semula}&\\ x^{2}-10x-37&=2\\ x^{2}-10x-39&=0\\ (x-13)(x+3)&=0\\ x=13\: \: \textrm{atau}\: \: x=-3& \end{aligned}\\ &\textrm{Jadi},\: \: \color{red}x=13 \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Diketahui akar-akar persamaan kuadrat}\\ &x^{2}+x-3=0\: \: \textrm{adalah}\: \: \alpha \: \: \textrm{dan}\: \beta .\: \textrm{Tentukanlah nilai dari}\\ &\alpha ^{3}-4\beta ^{2}+19\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &\begin{array}{|l|l|l|}\hline x^{2}+x-3=0&\begin{aligned}\alpha ^{2}+\alpha -3&=0\\ &\\ \Leftrightarrow \alpha ^{2}&=3-\alpha.....(1) \end{aligned}&\begin{aligned}\beta ^{2}+\beta -3&=0\\ &\\ \Leftrightarrow \beta ^{2}&=3-\beta.....(2) \end{aligned}\\\hline \left\{\begin{matrix} \alpha +\beta =\displaystyle \frac{-b}{a}=-1\\ \alpha \beta =\displaystyle \frac{c}{a}=-3 \end{matrix}\right.&\begin{aligned}\alpha ^{3}+\alpha ^{2}-3\alpha &=0\\ &\\ \Leftrightarrow \alpha ^{3}&=3\alpha -\alpha ^{2}.....(3) \end{aligned}&\begin{aligned}\beta ^{3}+\beta ^{2}-3\beta &=0\\ &\\ \Leftrightarrow \beta ^{3}&=3\beta -\beta ^{2} .....(4)\end{aligned}\\\hline \end{array}\\ &\\ &\begin{aligned}\alpha ^{3}-4\beta ^{2}+19&=\left ( 3\alpha -\alpha ^{2} \right )-4\left ( 3-\beta \right )+19,\: \textnormal{perhatikan persamaan}\: \: (3)\: \: \textrm{dan}\: \: (2)\\ &=3\alpha -\left ( 3-\alpha \right )-12+4\beta +19\\ &=4\alpha +4\beta -3+7\\ &=4\left ( \alpha +\beta \right )+4\\ &=4(-1)+4\\ &=0 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Akar real terbesar untuk persamaan}\\ &\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^{2}-11x-4\\ &\textrm{adalah}\: \: p+\sqrt{q+\sqrt{r}},\: \textrm{dengan}\: \: p,\: q,\: \textrm{dan}\: \: r\\ &\textrm{adalah bilangan-bilangan asli}.\: \textrm{Carilah hasil}\: \: p+q+r\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}&=x^{2}-11x-4\\ \frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1&=x^{2}-11x\\ \frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19}&=x^{2}-11x\\ \frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}&=x^{2}-11x\\ \frac{x(x-19)+x(x-3)}{(x-3)(x-19)}+\frac{x(x-17)+x(x-5)}{(x-5)(x-17)}&=x^{2}-11x\\ \frac{2x^{2}-22x}{x^{2}-22x+57}+\frac{2x^{2}-22}{x^{2}-22x+85}&=x^{2}-11x\\ \left ( x^{2}-11x \right )\left ( \frac{2}{x^{2}-22x+57}+\frac{2}{x^{2}-22x+85} \right )&=x^{2}-11x,\quad \textrm{misal}\: \: t=x^{2}-22x\\ \left ( \frac{2}{t+57}+\frac{2}{t+85} \right )&=\frac{x^{2}-11x}{x^{2}-11x}=1\\ 2\left ( t+85 \right )+2\left ( t+57 \right )&=(t+57)(t+85)\\ 2t+170+2t+114&=t^{2}+142t+4845\\ 0&=t^{2}+138t+4731\\ t^{2}+138t+4731&=0\: \: \left\{\begin{matrix} a=1\\ b=138\\ c=4731 \end{matrix}\right.\\ t_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ t_{1,2}&=\displaystyle \frac{-138\pm \sqrt{138^{2}-4.1.4731}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{19044-18924}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{120}}{2}\\ &=\displaystyle \frac{-138\pm 2\sqrt{30}}{2}\\ &=-69\pm \sqrt{30} \end{aligned}\\ &\begin{aligned}\textrm{Selanjutnya}\qquad\qquad\qquad\qquad\qquad &\\ t_{1,2}&=-69\pm \sqrt{30}\\ x^{2}-22x&=-69\pm \sqrt{30}\\ x^{2}-22x+69\pm \sqrt{30}&=0\\ x^{2}-22x+69+\sqrt{30}&=0\quad \textrm{atau}\quad x^{2}-22x+69-\sqrt{30}\\ &\\ \textrm{dengan cara yang} &\: \: \textrm{semisal diatas}\\  &\\ x_{1,2}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69+\sqrt{30} \right )}}{2}&\qquad \textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69-\sqrt{30} \right )}}{2}\\ x_{1,2}=\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}&\qquad \textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{484-276+4\sqrt{30}}}{2}\\ x_{1,2}=\displaystyle \frac{22\pm \sqrt{208-4\sqrt{30}}}{2}&\qquad \textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{208+4\sqrt{30}}}{2}\\ x_{1,2}=\displaystyle \frac{22\pm 2\sqrt{52-\sqrt{30}}}{2}&\qquad \textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}\\ x_{1,2}=11\pm \sqrt{52-\sqrt{30}}&\qquad \textrm{atau}\qquad x_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ &\\  \textrm{Maka},\qquad\qquad\quad &\\ \left\{\begin{matrix} x_{1}=11+\sqrt{52-\sqrt{30}}\\ \\ x_{2}=11-\sqrt{52-\sqrt{30}} \end{matrix}\right.&\qquad \textrm{atau}\qquad \left\{\begin{matrix} x_{3}=11+\sqrt{52+\sqrt{30}}\\ \\ x_{4}=11-\sqrt{52+\sqrt{30}} \end{matrix}\right. \end{aligned}\\\\\\ &\begin{aligned}\textrm{Selanjutnya nilai}&\: \textrm{yang paling pas sesuai soal adalah}\: x_{3}=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \textrm{Sehingga nilai}\: \: \: \: \, \, &p+q+r=11+52+30=93 \end{aligned} \end{array}$

Sumber Referensi:

1. Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.
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