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

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

$\begin{array}{l}\\ 17.& \text{Batas nilai}m\text{yang menyebabkan fungsi kuadrat}\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{definit positif adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle m>-2}& & & & & \\ \text{B}.& {\displaystyle m>1}& & \\ \text{C}.& m<1\\ \text{D}.& -2<m<1\\ \text{E}.& m<-{\displaystyle 2\text{atau}m>1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{Syarat fungsi kuadrat definit positif}:\\ & \bullet a>0\\ & \bullet D=b^2-4ac<0\\ & \text{maka}\\ & a=m-1>0\iff m>1,\text{dan}\\ & D=(2(m-1){)}^2-4(m-1)(2m+1)<0\\ & \iff 4(m-1{)}^2-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4)-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4-8m-4)<0\\ & \iff (m-1)(-4m-8)<0,\mathbf{\text{dibagi}}-4\\ & \iff (m-1)(m+2)>0\end{array}\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle 2}& & & & 1& & \mathbf{\text{Sumbu}}-m\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ m=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}$.

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

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

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

$\begin{array}{l}\\ 20.& \text{Fungsi}f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{akan memotong sumbu-X di titik}(1,0)\text{dan}\\ & \text{memenuhi}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle \frac{b(c-a)}{a(b-c)}}}& & & & & \text{D}.& {\displaystyle {\displaystyle \frac{c(a-b)}{a(b-c)}}}\\ \\ \text{B}.& {\displaystyle {\displaystyle \frac{a(b-c)}{c(a-b)}}}& & \text{C}.& {\displaystyle {\displaystyle \frac{a(b-c)}{b(c-a)}}}& & \text{E}.& {\displaystyle {\displaystyle \frac{c(a-b)}{b(c-a)}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui bahwa FK}:\\ & f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{maka nilai}\\ & \bullet x_1+x_2=-{\displaystyle \frac{b(c-a)}{a(b-c)}={\displaystyle \frac{b(a-c)}{a(b-c)}}}\\ & \bullet x_1\times x_2={\displaystyle \frac{c(a-b)}{a(b-c)}}\end{array}$

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

 $\begin{array}{l}\\ 11.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{pada gambar tersebut di atas dengan}\\ & \text{titik balik}(-2,{\displaystyle \frac{9}{2}})\text{dan melalui}(-1,4)\\ & \text{adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y={\displaystyle \frac{1}{2}(2+4x-x^2)}}& & & & & \\ \text{B}.& {\displaystyle y={\displaystyle \frac{1}{2}(5-4x-x^2)}}& & \\ \text{C}.& y={\displaystyle \frac{1}{2}(5-2x-x^2)}\\ \text{D}.& y=1-4x-x^2\\ \text{E}.& y=5+3x-x^2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(-2,{\displaystyle \frac{9}{2}})\text{dan melalui}\\ & \text{titik}(-1,4),(-5,0),\text{serta}(1,0),\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}=a(x-x_1)(x-x_2)\\ & \text{Pilih salah satunya, di sini saya pilih formula}\\ & \text{yang kedua, yaitu}:y=a(x-x_1)(x-x_2)\\ & \iff 4=a(-1-(-5))(-1-1)\iff 4=a(4)(-2)\\ & \iff a=-{\displaystyle \frac{1}{2}}\\ & \text{Selanjutnya kembalikan ke formula semula}\\ & \text{yaitu}:\\ & y=a(x-x_1)(x-x_2)=-{\displaystyle \frac{1}{2}(x-(-5))(x-1)}\\ & \iff y=-{\displaystyle \frac{1}{2}(x+5)(x-1)={\displaystyle \frac{1}{2}(5-4x-x^2)}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Gambar berikut adalah fungsi parabola}\\ & \text{dengan persamaan}y=a{x}^2-4x+k\end{array}$.

$.\begin{array}{l}\\ & \text{Jika nilai minimum}y\text{adalah}-8\text{maka}\\ & \text{nilai}k\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -5}& & & & & \text{D}.& {\displaystyle 2}\\ \text{B}.& {\displaystyle -4}& & \text{C}.& {\displaystyle -2}& & \text{E}.& 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2-4x+k,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(3,-8)\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}\\ & -2=a(0-3{)}^2+(-8)\iff -2+8=9.a\\ & 6=9a\iff a={\displaystyle \frac{6}{9}=\frac{2}{3}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y={\displaystyle \frac{2}{3}(x-3{)}^2-8=\frac{2}{3}(x^2-6x+9)-8}\\ & \iff y={\displaystyle \frac{2}{3}{x}^2-4x+6-8={\displaystyle \frac{2}{3}{x}^2-4x-2}}\\ & \text{Jadi, nilai}k=-2\end{array}\end{array}$.

 $\begin{array}{l}\\ 13.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{gambar di atas adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y=-(x-3{)}^2-1}& & & & & \\ \text{B}.& {\displaystyle y=-(x-3{)}^2+1}& & \\ \text{C}.& y=-{\displaystyle \frac{1}{3}{(x-3)}^2+1}\\ \text{D}.& y=-{\displaystyle \frac{1}{3}{(x-3)}^2-1}\\ \text{E}.& y=-{\displaystyle \frac{1}{3}{(x+3)}^2-1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK dengan koordinat}\\ & (x_{ss},y_{ss})=(3,-1)\text{dan melalui titik}(0,-4)\\ & \text{maka}y=a(x-x_{ss}{)}^2+y_{ss}\\ & -4=a(0-3{)}^2+(-1)\iff -4+1=9.a\\ & -3=9a\iff a=-{\displaystyle \frac{1}{3}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y=-{\displaystyle \frac{1}{3}(x-3{)}^2-1}\end{array}\end{array}$.

 $\begin{array}{l}\\ 14.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika grafik fungsi kuadrat di atas adalah}\\ & y=a{x}^2+bx+c,\text{maka hasil kali dari}\\ & a.b.c\text{ adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -20}& & & & & \text{D}.& {\displaystyle 3}\\ \text{B}.& {\displaystyle -6}& & \text{C}.& {\displaystyle -3}& & \text{E}.& 20\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:a{x}^2+bx+c\\ & (x_{ss},y_{ss})=(-2,-4)\text{dan melalui titik}(0,-6)\\ & \text{maka}y=a(x-x_{ss}{)}^2+y_{ss}\\ & -6=a(0-(-4){)}^2+(-2)\iff -6+2=16.a\\ & -4=16a\iff a=-{\displaystyle \frac{1}{4}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y=-{\displaystyle \frac{1}{4}(x+4{)}^2-2}\\ & =-{\displaystyle \frac{1}{4}{x}^2-2x-6\{\begin{array}{c}a=-{\displaystyle \frac{1}{4}}\\ b=-2\\ c=-6\end{array}}\\ & \text{Sehingga nilai}abc=(-{\displaystyle \frac{1}{4}}).(-2).(-6)=-3\end{array}\end{array}$.

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

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

 $\begin{array}{l}\\ 6.& \text{Fungsi kuadrat}f(x)=(2x+p{)}^2+q\\ & \text{dengan titik balik minimum}(-1,3).\\ & \text{Nilai}p+q\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 2}& & & & & \text{D}.& {\displaystyle 6}\\ \text{B}.& {\displaystyle 4}& & \text{C}.& {\displaystyle 5}& & \text{E}.& 7\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=(2x+p{)}^2+q,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-1,3),\text{maka}\\ & f(x)=4x^2+4px+p^2+q\text{dengan}\\ & x_{ss}=-1=-{\displaystyle \frac{b}{2a}\iff 1={\displaystyle \frac{4p}{2.4}\iff p=2}}\\ & \text{Selanjutnya}\\ & f(-1)=(2.(-1)+2{)}^2+q=3\iff q=3\\ & \text{maka}\\ & p+q=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika grafik fungsi kuadrat}f(x)=a{x}^2+x+c\\ & \text{dengan titik balik minimum}(-1,3)\text{dan melalui}\\ & (2,12)\text{maka}a+b+c\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 7}& & & & & \text{D}.& {\displaystyle 13}\\ \text{B}.& {\displaystyle 9}& & \text{C}.& {\displaystyle 11}& & \text{E}.& 15\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-1,3)\text{dan melalui titik}\\ & (2,12),\text{maka}\\ & \overline{\begin{array}{cc}\begin{array}{rl}& 12=4a+2b+c\\ & 3=a-b+c\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-\\ & 9=3a+3b\\ & \iff 3=a+b\end{array}& \begin{array}{rl}& -{\displaystyle \frac{b}{2a}=-1,\text{maka}}\\ & 2a-b=0,\text{dan ingat}\\ & a+b=3\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}+\\ & 3a=3\\ & \iff a=1,\text{maka}b=2\end{array}\end{array}}\\ & \text{dan}\\ & a-b+c=3\Rightarrow 1-2+c=3\Rightarrow c=4\\ & \text{Jadi, nilai}a+b+c=1+2+4=7\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai minimum grafik fungsi}f(x)=a{x}^2-2x+8\\ & \text{adalah 5. Nilai }6a\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 1}& & & & & \text{D}.& {\displaystyle 9}\\ \text{B}.& {\displaystyle 2}& & \text{C}.& {\displaystyle 4}& & \text{E}.& 12\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2-2x+8,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-{\displaystyle \frac{b}{2a},5})\text{maka}\\ & \bullet x_{ss}=-{\displaystyle \frac{b}{2a}=-\frac{-2}{2a}=\frac{1}{a}}\\ & \bullet y_{ss}=f\left(x_{ss}\right)=a{\left({\displaystyle \frac{1}{a}}\right)}^2-2\left({\displaystyle \frac{1}{a}}\right)+8=5\\ & \iff {\displaystyle \frac{1}{a}-\frac{2}{a}=5-8\iff -{\displaystyle \frac{1}{a}=-3}}\\ & \iff a={\displaystyle \frac{1}{3}}\\ & \text{maka nilai}6a=6\left({\displaystyle \frac{1}{3}}\right)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika kurva fungsi}f(x)=x^2+bx+c\\ & \text{memotong sumbu-X di}(1,0)\text{dan}(5,0),\\ & \text{maka nilai}{b}^2-c^2\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -11}& & & & & \text{D}.& {\displaystyle 11}\\ \text{B}.& {\displaystyle -3}& & \text{C}.& {\displaystyle 6}& & \text{E}.& 13\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=x^2+bx+c,\text{memotong}\\ & \text{sumbu-X di}(1,0)\mathrm{\&}(5,0)\text{artinya}{x}_1=1\mathrm{\&}{x}_2=5\\ & \text{maka}\\ & x_{ss}={\displaystyle \frac{-b}{2.1}={\displaystyle \frac{x_1+x_2}{2}=\frac{1+5}{2}\iff b=-6}}\\ & \text{dan kita juga memiliki}\\ & f(1)=1+b+c=0\Rightarrow c=-b-1=6-1=5\\ & \text{Sehingga}\\ & b^2-c^2=(-6{)}^2-5^2=36-25=11\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{pada gambar tersebut di atas adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y=6x^2-12x+18}& & & & & \\ \text{B}.& {\displaystyle y=6x^2+12x+16}& & \\ \text{C}.& y=6x^2-24x+17\\ \text{D}.& y=6x^2-24x+19\\ \text{E}.& y=6x^2-24x+29\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(2,-5)\text{dan melalui}\\ & \text{titik}(3,1),\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}\iff 1=a(3-2{)}^2+(-5)\\ & \iff 6=a{.1}^2\iff a=6\\ & \text{maka persamaan fungsi kuadratnya adalah}:\\ & y=a(x-x_{ss}{)}^2+y_{ss}\\ & \iff y=6(x-2{)}^2-5\iff y=6(x^2-4x+4)-5\\ & \iff y=6x^2-24x+19\end{array}\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(x)=x^2-2x-15.\text{Jika}\\ & \text{domain}\{x|-4\le x\le 2,x\in \mathbb{R}\},\text{maka}\\ & range\text{-nya adalah}....\\ & \begin{array}{lllll}\text{A}.& -15\le f(x)\le 20& & & \\ \text{B}.& -15\le f(x)\le 9& & \\ \text{C}.& {\displaystyle -16\le f(x)\le 9}& & \\ \text{D}.& {\displaystyle -16\le f(x)\le 20}& & \\ \text{E}.& -15\le f(x)\le 5& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=x^2-2x-15,\text{dengan}\\ & {\text{D}}_f=\{x|-4\le x\le 2,x\in \mathbb{R}\},\\ & \text{maka}range\text{fungsinya adalah}{\text{R}}_f,\text{di mana}\\ & {\text{R}}_f\text{diperoleh dengan cara di antaranya}\\ & \text{mensubstitusikan langsung ke fungsinya, yaitu}:\\ & f(-4)=(-4{)}^2-2(-4)-15=9\\ & f(-3)=(-3{)}^2-2(-3)-15=0\\ & f(-2)=(-2{)}^2-2(-2)-15=-7\\ & f(-1)=(-1{)}^2-2(-1)-15=-12\\ & f(0)=(0{)}^2-2(0)-15=-15\\ & f(1)=(1{)}^2-2(1)-15=-16\\ & f(2)=(2{)}^2-2(2)-15=-15\\ & \text{Jadi, range fungsinya}:{\text{R}}_f={\displaystyle -16\le f(x)\le 9}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Daerah hasil fungsi}f(x)=-x^2+6x-5\text{untuk}\\ & \text{daerah asal}\{x|-1\le x\le 6,x\in \mathbb{R}\}\text{dan}\\ & y=f(x)\text{adalah}....\\ & \begin{array}{lllll}\text{A}.& \{y|-5\le y\le 0,y\in \mathbb{R}\}& & & \\ \text{B}.& \{y|-12\le y\le 4,y\in \mathbb{R}\}& & \\ \text{C}.& {\displaystyle \{y|-4\le y\le 1,y\in \mathbb{R}\}}& & \\ \text{D}.& {\displaystyle \{y|-5\le y\le 4,y\in \mathbb{R}\}}& & \\ \text{E}.& \{y|-1\le y\le 6,y\in \mathbb{R}\}& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Masih sama dengan cara di atas. Diketahui FK}:\\ & f(x)=-x^2+6x-5,\text{dengan}\\ & {\text{D}}_f=\{x|-1\le x\le 6,x\in \mathbb{R}\},\\ & \text{maka}range\text{fungsinya adalah}{\text{R}}_f,\text{di mana}\\ & {\text{R}}_f\text{diperoleh dengan cara di antaranya}\\ & \text{mensubstitusikan langsung ke fungsinya, yaitu}:\\ & f(-1)=-(-1{)}^2+6(-1)-5=-12\\ & f(0)=-(0{)}^2+6(0)-5=-5\\ & f(1)=-(1{)}^2+6(1)-5=0\\ & f(2)=-(2{)}^2+6(2)-5=3\\ & f(3)=-(3{)}^2+6(3)-5=4\\ & f(4)=-(4{)}^2+6(1)-5=3\\ & f(5)=-(5{)}^2+6(5)-5=0\\ & f(6)=-(6{)}^2+6(6)-5=-5\\ & \text{Jadi, range fungsinya}:{\text{R}}_f=\{y|-12\le y\le 4,y\in \mathbb{R}\}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 4.& \text{Fungsi kuadrat dengan titik balik minimum}\\ & (3,-4)\text{dan melalui titik}(0,5)\text{adalah}....\\ & \begin{array}{lllll}\text{A}.& y=x^2-6x+5& & & \\ \text{B}.& y=x^2+6x+5& & \\ \text{C}.& {\displaystyle y=2x^2-6x+5}& & \\ \text{D}.& {\displaystyle y=2x^2+6x+5}& & \\ \text{E}.& y=2x^2-6x-5& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:y=f(x)=a{(x-x_{ss})}^2+y_{ss}\\ & \text{Koordinat titik baliknya}=(x_{ss},y_{ss})=(3,-4)\\ & \text{dan melalui titik}(0,5),\text{maka}\\ & 5=a(0-3{)}^2+(-4)\iff 5+4=a.9\iff a={\displaystyle \frac{9}{9}=1}\\ & \text{Sehingga Fk-nya dengan}a=1\text{adalah}:\\ & f(x)=a{(x-x_{ss})}^2+y_{ss}=1.(x-3{)}^2+(-4)\\ & =(x^2-6x+9)-4\\ & =x^2-6x+5\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Fungsi kuadrat yang melalui titik}(0,2)\text{dan}\\ & (-1,0)\text{dengan sumbu simetri garis}\\ & x={\displaystyle \frac{1}{2}\text{adalah}....}\\ & \begin{array}{lllll}\text{A}.& y=(x+1)(2-x)& & & \\ \text{B}.& y=(x-1)(x+2)& & \\ \text{C}.& {\displaystyle y=2-x-x^2}& & \\ \text{D}.& {\displaystyle y=x^2-x+2}& & \\ \text{E}.& y=-(x-1)(x+2)& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:y=f(x)=a{(x-x_{ss})}^2+y_{ss}\\ & \text{atau}y=f(x)=a(x-x_1)(x-x_2)\text{dengan}\\ & x_1\text{dan}{x}_2\text{sebagai akar-akarnya}\\ & \text{Dan diketahui pula sebagaimana keterangan}\\ & \text{dalam soal, maka},x_1=-1,x_{ss}={\displaystyle \frac{1}{2}}\\ & \text{Sehingga}\\ & x_{ss}=-{\displaystyle \frac{b}{2a}={\displaystyle \frac{x_1+x_2}{2}\iff {\displaystyle \frac{1}{2}=\frac{-1+x_2}{2}\iff x_2=2}}}\\ & \text{Selanjutnya garfik melalui}(0,2),\text{maka}\\ & y=a(x-x_1)(x-x_2)\iff 2=a(0-(-1))(0-2)\\ & \iff 2=a(1)(-2)\iff a=-1\\ & \text{Sehingga fungsi akan berupa}\\ & f(x)=a(x-x_1)(x-x_2)=-1(x+1)(x-2)\\ & =(x+1)(2-x)\end{array}\end{array}$.

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

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

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

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

$\begin{array}{l}\\ 12.& \text{Jika}{x}^2+2\left|x\right|-8=0,\text{maka nilai}x\\ & \text{yang memenuhi adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -4}& & & & & \text{D}.& {\displaystyle 0}\\ \text{B}.& {\displaystyle -2}& & \text{C}.& {\displaystyle -1}& & \text{E}.& 4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & x^2+2\left|x\right|-8=0\iff (\left|x\right|+4)(\left|x\right|-2)=0\\ & \iff \left|x\right|=-4(\text{bukan solusi})\text{atau}\left|x\right|=2(\text{solusi})\\ & \text{Pilih}\left|x\right|=2\Rightarrow x=\pm 2\end{array}$.

$\begin{array}{l}\\ 13.& \text{Jika}\alpha \text{dan}\beta \text{akar-akar dari persamaan}\\ & x^2-2x=|x-1|+5,\text{maka nilai }\\ & \alpha +\beta \text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -2}& & & & & \text{D}.& {\displaystyle 1}\\ \text{B}.& {\displaystyle -1}& & \text{C}.& {\displaystyle 0}& & \text{E}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & \begin{array}{rl}& x^2-2x=|x-1|+5\iff x^2-2x-5=|x-1|\\ & \text{Untuk}x\ge 1,\text{persamaan akan menjadi}\\ & x^2-2x-5=x-1\iff x^2-2x-x-5+1=0\\ & x^2-3x-4=0\iff (x-4)(x+1)=0\\ & \iff x=4(\text{memenuhi})\text{atau}x=-1(\text{tidak})\\ & \text{Untuk}x<1,\text{persamaan akan menjadi}\\ & x^2-2x-5=1-x\iff x^2-2x+x-5-1=0\\ & x^2-x-6=0\iff (x-3)(x+2)=0\\ & \iff x=3(\text{tidak})\text{atau}x=-2(\text{memenuhi})\\ & \text{Pilih}\alpha =4,\text{dan}\beta =-2,\text{maka}\\ & \alpha +\beta =4+(-2)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Persamaan kuadrat}{x}^2-2x+m=0\text{mempunyai }\\ & \text{akar-akar yang rasional, maka nilai}m\text{yang}\\ & \text{mungkin adalah}....\\ & \begin{array}{lllllll}\text{A}.& 1-{\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & & \\ \text{B}.& 1+{\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{C}.& {\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{D}.& {\displaystyle \frac{k^2-1}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{E}.& 1-{\displaystyle \frac{k}{4}}& \text{untuk}& k=0,1,2,\cdots & & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Akar-akar dari PK}:x^2-2x+m=0\\ & x_{1,2}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{2\pm \sqrt{4-4m}}{2}}\\ & \text{Agar nilai}m\text{rasional, maka}\\ & 4-4m=k^2\iff 4m=4-k^2\iff m=1-{\displaystyle \frac{k^2}{4}}\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 6.& \text{Diketahui}{x}_1\text{dan}{x}_2\text{akar-akar dari persamaan}\\ & x^2-3x-4=0.\text{Persamaan kuadrat baru yang}\\ & \text{memiliki akar-akar}2x_1\text{dan}2x_2\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 2x^2+6x-16=0}& & & & & \text{D}.& {\displaystyle x^2-6x-16=0}\\ \text{B}.& {\displaystyle 2x^2-6x-16=0}& & & & & \text{E}.& {\displaystyle x^2+6x-16=0}\\ \text{C}.& x^2+6x+16=0& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui bahwa PK}:x^2-3x-4=0\\ & \text{dengan}a=1,b=-3,\text{dan}c=-4\\ & \text{Alternatif 1}\\ & \text{Misalkan Persamaan Kuadrat baru dengan}\\ & \text{akar-akar}\alpha =2x_1\text{dan}\beta =2x_2,\text{adalah}\\ & x^2-(\alpha +\beta )x+\alpha \beta =0\\ & \iff x^2-(2x_1+2x_2)x+2x_1\times 2x_2=0\\ & \iff x^2-2(x_1+x_2)x+4x_1{x}_2=0\\ & \iff x^2-2\left({\displaystyle -\frac{b}{a}}\right)x+4\left({\displaystyle \frac{c}{a}}\right)=0\\ & \iff x^2-2(3)x+4(-4)=0\iff x^2-6x-16=0\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& \text{PK lama}:x^2-3x-4=0\\ & \text{dengan}{x}_1\text{dan}{x}_2\\ & \text{PK baru dengan}2x_1\text{dan}2x_2\\ & \text{PK baru}:x^2-3(2)x-4(2^2)=0\\ & \iff x^2-6x-16=0\\ & \text{Formula tersebut dapat digunakan},\\ & \text{syaratnya koefisien dari}{x}^2=1\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Diketahui}{x}_1\text{dan}{x}_2\text{akar-akar dari persamaan}\\ & 2x^2-3x+4=0.\text{Persamaan kuadrat baru yang}\\ & \text{memiliki akar-akar}2x_1-1\text{dan}2x_2-1\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 2x^2+x-6=0}& & & & & \text{D}.& {\displaystyle x^2+x-6=0}\\ \text{B}.& {\displaystyle x^2+5x+6=0}& & & & & \text{E}.& {\displaystyle x^2-x+6=0}\\ \text{C}.& x^2-5x+6=0& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui bahwa PK}:2x^2-3x+4=0\\ & \text{dengan}a=2,b=-3,\text{dan}c=4\\ & \begin{array}{rl}& \text{Misalkan Persamaan Kuadrat baru dengan}\\ & \text{akar-akar}\alpha =2x_1-1\text{dan}\beta =2x_2-1,\text{adalah}\\ & x^2-(\alpha +\beta )x+\alpha \beta =0\\ & \iff x^2-(2x_1-1+2x_2-1)x+(2x_1-1)\times (2x_2-1)=0\\ & \iff x^2-(2(x_1+x_2)-2)x+4x_1{x}_2-2(x_1+x_2)+1=0\\ & \iff x^2-(2\left({\displaystyle -\frac{b}{a}}\right)-2)x+4\left({\displaystyle \frac{c}{a}}\right)-2(-{\displaystyle \frac{b}{a}})+1=0\\ & \iff x^2-(2(3/2)-2)x+4(4/2)-2(3/2)+1=0\\ & \iff x^2-x+(8-3+1)=0\iff x^2-x+6=0\end{array}\end{array}$.

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

$\begin{array}{l}\\ 9.& \text{Akar-akar dari persamaan}{x}^2+(m-1)x-5=0\\ & \text{adalah}{x}_1\text{dan}{x}_2.\text{Jika}{x}_1^2+x_2^2-2x_1{x}_2=8m,\\ & \text{maka nilai}m\text{ adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -6\text{atau}-14}& & & & & \text{D}.& {\displaystyle 3\text{atau}7}\\ \text{B}.& {\displaystyle 6\text{atau}14}& & & & & \text{E}.& {\displaystyle -3\text{atau}-7}\\ \text{C}.& 3\text{atau}-7& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui}{x}^2+(m-1)x-5=0\\ & \text{dengan akar-akar}{x}_1\text{dan}{x}_2\{\begin{array}{c}{x}_1+x_2=1-m\\ x_1\times x_2=-5\end{array}\\ & \text{Selanjutnya},\\ & x_1^2+x_2^2-2x_1{x}_2=8m\\ & \iff {(x_1+x_2)}^2-4x_1{x}_2=8\iff (1-m{)}^2-4(-5)=8m\\ & \iff 1-2m+m^{}+20-8m=0\\ & \iff m^2-10m+21=0\iff (m-3)(m-7)=0\\ & \iff m=3ataum=7\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 1.& \text{Penyelesaian terkecil dari persamaan kuadrat}\\ & (x-{\displaystyle \frac{3}{4}})(x-{\displaystyle \frac{3}{4}})+(x-{\displaystyle \frac{3}{4}})(x-{\displaystyle \frac{1}{2}})=0\\ & \text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& -{\displaystyle \frac{3}{4}}& & & & & \text{D}.& {\displaystyle \frac{3}{4}}\\ \text{B}.& {\displaystyle \frac{1}{2}}& & \text{C}.& {\displaystyle \frac{5}{8}}& & \text{E}.& 1\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{3}{4}})+\left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{1}{2}})=0\\ & \iff \left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{3}{4}+x-\frac{1}{2}})=0\\ & \iff \left(x-{\displaystyle \frac{3}{4}}\right)(2x-{\displaystyle \frac{5}{4}})=0\\ & \iff x-{\displaystyle \frac{3}{4}=0\text{atau}2x-\frac{5}{4}=0}\\ & \iff x={\displaystyle \frac{3}{4}\text{atau}2x=\frac{5}{4}}\\ & \iff x={\displaystyle \frac{6}{8}\text{atau}x=\frac{5}{8}}\end{array}\\ & \text{Jadi, nilai terkecilnya adalah}{\displaystyle \frac{5}{8}}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}m\text{dan}n\text{adalah penyelesaian dari}\\ & \text{persamaan kuadrat}{x}^2+mx+n=0,\\ & \text{dengan}m\ne 0\text{dan}n\ne 0\text{jumlah kedua}\\ & \text{penyelesaian tersebut adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -\frac{1}{2}}& & & & & \text{D}.& {\displaystyle 1}\\ \text{B}.& {\displaystyle -1}& & \text{C}.& {\displaystyle \frac{1}{2}}& & \text{E}.& {\displaystyle \text{tidak dapat ditentukan}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \text{Diketahui bahwa PK}:x^2+mx+n=0\\ & \text{dengan}a=1,b=m,\text{dan}c=n\\ & \begin{array}{rl}& \bullet x_1+x_2=-{\displaystyle \frac{b}{a}\iff m+n=-\frac{m}{1}}\\ & \bullet x_1\times x_2={\displaystyle \frac{c}{a}\iff mn=\frac{n}{1}\iff m=1}\\ & \mathbf{\text{Dari persamaan pertama akan diperoleh}}\\ & m+n=-m=-1\end{array}\\ & \text{Jadi, nilai m+n adalah}{\displaystyle -1}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika persamaan}2x^2-3x-14=0\text{mempunyai}\\ & \text{akar-akar}{x}_1\text{dan}{x}_2\text{dengan}{x}_1>x_2,\text{maka}\\ & 2x_1+3x_2\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 2}& & & & & \text{D}.& {\displaystyle -2}\\ \text{B}.& {\displaystyle 1}& & \text{C}.& {\displaystyle -1}& & \text{E}.& -5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \text{Diketahui bahwa PK}:2x^2-3x-14=0\\ & \begin{array}{rl}& 2x^2-3x-14=0\iff {\displaystyle \frac{(2x-7)(2x+4)}{2}=0}\\ & \iff (2x-7)(x+2)=0\\ & \iff 2x-7=0\text{atau}x+2=0\\ & x={\displaystyle \frac{7}{2}\text{atau}x=-2}\\ & \text{Karena nilai dari}{x}_1>x_2,\text{maka}\\ & x_1={\displaystyle \frac{7}{2}\text{dan}{x}_2=-2}\end{array}\\ & \text{Sehingga nilai}2x_1+3x_2=2{\displaystyle \frac{7}{2}+3(-2)=7-6=1}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}{x}_1\text{dan}{x}_2\text{adalah akar-akar dari}\\ & \text{persamaan kuadrat}{x}^2+6x+2=0,\\ & \text{nilai dari}{x}_1^2+x_2^2-4x_1{x}_2\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 28}& & & & & \text{D}.& {\displaystyle 18}\\ \text{B}.& {\displaystyle 26}& & \text{C}.& {\displaystyle 24}& & \text{E}.& {\displaystyle 16}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \text{Diketahui bahwa PK}:x^2+6x+2=0\\ & \text{dengan}a=1,b=6,\text{dan}c=2\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& x_1^2+x_2^2-4x_1{x}_2\\ & ={(x_1+x_2)}^2-2x_1{x}_2-4x_1{x}_2\\ & ={(x_1+x_2)}^2-6x_1{x}_2\\ & ={(-{\displaystyle \frac{b}{a}})}^2-6\left({\displaystyle \frac{c}{a}}\right)=(-6{)}^2-6(2)\\ & =36-12\\ & =24\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& x^2+6x+2=0\iff x^2=-6x-2\\ & \bullet x=x_1\Rightarrow x_1^2=-6x_1-2\\ & \bullet x=x_2\Rightarrow x_2^2=-6x_2-2\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}+\\ & \iff x_1^2+x_2^2=-6(x_1+x_2)-4\\ & \iff x_1^2+x_2^2-4x_1{x}_2=-6(x_1+x_2)-4x_1{x}_2-4\\ & =-6(-{\displaystyle \frac{b}{a}})-4\left(\frac{c}{a}\right)-4\\ & =-6(-6)-4(2)-4\\ & =36-8-4\\ & =24\end{array}\\ & \text{Jadi, nilai yang dimaksud adalah}{\displaystyle 24}\end{array}$.

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

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

$\begin{array}{|ll|}\hline \text{Pengertian}& \begin{array}{rl}& \text{Suatu fungsi yang berbentuk}\\ & f(x)=a{x}^2+bx+c\\ & a,b,c,\in \mathbb{R},a\ne 0\end{array}\\ \text{Grafik Fungsi}& \text{Keterangan}\\ \text{Titik potong sumbu x}& \text{Jika ada}\\ & \begin{array}{rl}& \text{untuk titik potong}\\ & \text{terhadap sumbu x }\\ & \text{Jika y = 0 maka }\\ & a{x}^2+bx+c=0\\ & \text{Selanjutnya tinggal}\\ & \text{menentukan nilai D}\\ & D=b^2-4ac\text{adalah}\\ & \text{nilai diskriminan}.\\ & \text{Jika}D>0\\ & \text{maka grafik}\\ & \text{memotong sumbu x}\\ & \text{di dua tempat berbeda}\\ & \text{yaitu di}(x_1,0)\text{dan}(x_2,0).\\ & \text{dan jika D = 0}\\ & \text{maka grafik}\\ & \text{ hanya menyinggung}\\ & \text{sumbu x di satu titik}\\ & \text{yaitu di }(x_1,0)\\ & \text{dan jika}D<0\\ & \text{maka grafik}\\ & \text{tidak memotong}\\ & \text{atau menyinggung sumbu x}\end{array}\\ \text{Titik potong sumbu y}& \begin{array}{rl}& \text{titik potong terhadap}\\ & \text{sumbu y, jika x = 0}\\ & y=f(x)=a{x}^2+bx+c\\ & y=f(0)=a(0{)}^2+b(0)+c\\ & y=c\end{array}\\ \text{Sumbu Simetri (SS)}& x={\displaystyle \frac{-b}{2a}}\\ \text{Titik Puncak}& \left({\displaystyle \frac{-b}{2a},{\displaystyle \frac{D}{-4a}}}\right)\\ \text{Posisi grafik}& \text{Jika}a>0\text{maka}\\ & \text{grafik terbuka ke atas}\\ & \text{Dan jika nilai}a<0\text{maka}\\ & \text{grafik terbuka ke bawah}\\ \hline\end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|cc|}\hline \text{Jika memotong sumbu}-\text{X}& \text{Jika menyinggung sumbu}-\text{X}\\ \text{di titik}(x_1,0)\text{dan}(x_2,0)& \text{di titik}(x_1,0)\text{dan melalui}\\ \text{dan melalui sebuah titik lain}& \text{sebuah titik lain}\\ & \\ y=f(x)=a(x-x_1)(x-x_2)& y=f(x)=a{(x-x_1)}^2\\ & \\ \text{Jika grafik fungsi itu melalui}& \text{Jika grafik fungsi itu melalui}\\ \text{Titik puncak}P(x_p,y_p)\text{dan}& \text{tiga buah titik yaitu}(x_1,y_1)\\ \text{sebuah titik lain}& (x_2,y_2)\text{dan}(x_3,y_3)\\ & \\ y=f(x)=a{(x-x_p)}^2+y_p& y=f(x)=a{x}^2+bx+c\\ & \\ \hline\end{array}$.

B. 3 Masalah yang Melibatkan Fungsi Kuadrat

$\begin{array}{rl}& y=f(x)=a{x}^2+bx+c\\ & \text{dengan}a,b,c\in \mathbb{R},a\ne 0\end{array}$.

B.3.1 Titik Stasioner

$\begin{array}{rl}& y_{ekstrim}=-{\displaystyle \frac{D}{4a}\Rightarrow \{\begin{array}{c}{y}_{\text{minimum}},\text{jika}a>0\\ y_{\text{maksimum}},\text{jika}a<0\end{array}}\\ & y_{\text{ekstrim}}\text{tercapai saat}x=-{\displaystyle \frac{b}{2a}}\\ & \mathbf{\text{Sehingga titik stasionernya adalah}}\\ & =(x_{ss},y_{ss})=(-{\displaystyle \frac{b}{2a},-{\displaystyle \frac{D}{4a}}})\end{array}$.

B.3.2 Definit Positif dan Definit Negatif

$\begin{array}{rl}& \text{Jika}D<0\text{dan}\{\begin{array}{c}a>0,\text{maka}y\text{akan selalu positif}\\ a<0,\text{maka}y\text{akan selalu negatif}\end{array}\\ & \text{untuk setiap nilai}x\end{array}$.

Perhatikan tambahan penjelasan berikut

$\begin{array}{rl}& \text{Tentang definit positif dan negatif}\\ & \begin{array}{c}\\ a>0.D<0& \text{Gambar}& \cup & \mathbf{\text{Sumbu-X}}\\ a<0,D<0& \text{Gambar}& \cap & \\ & \end{array}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f\text{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \text{maka nilai dari}f(4)-f(-2)\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=ax+b\\ & f(2)-f(-2)\\ & =(a(2)+b)-(a(-2)+b)=8\\ & 8=2a+2a\\ & 8=4a\\ & 2=a\\ & f(x)=2x+b,\text{dengan}b\text{konstan}\\ & \text{Sehingga nilai}\\ & f(4)-f(-2)=(2(4)+b)-(2(-2)+b)\\ & =8+b+4-b\\ & =12\end{array}\end{array}$.

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

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

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

$\begin{array}{rl}& {\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}\mathbf{=}\mathbf{0}\\ & \text{dengan}a,b,c\in \mathbb{R},a\ne 0\end{array}$.

Adapun cara penyelesaian persamaan kuadrat, jika $x_{1}dan{x}_2$ sebagai akar-akarya adalah:

$\begin{array}{|lll|}\hline \mathbf{\text{Pemfaktoran}}& \text{Melengkapkan}& \mathbf{\text{Rumus ABC}}\\ & \mathbf{\text{kuadrat sempurna}}& \\ (1)& (2)& (3)\\ \begin{array}{rl}& a{x}^2+bx+c=0\\ & (x-x_1)(x-x_2)=0\\ & \text{Jika koefisien}{x}^2\\ & \text{lebih dari 1, maka}\\ & \text{ubahlah ke bentuk}\\ & {\displaystyle \frac{1}{a}(ax-x_1)(ax-x_2)}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& a{x}^2+bx+c=0\\ & x^2+{\displaystyle \frac{b}{a}x+\frac{c}{a}=0}\\ & x^2+{\displaystyle \frac{b}{a}x=-\frac{c}{a}}\\ & \mathbf{\text{selanjutnya}}\\ & x^2+{\displaystyle \frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2}\\ & =-{\displaystyle \frac{c}{a}+{\left(\frac{b}{2a}\right)}^2}\\ & {(x+{\displaystyle \frac{b}{2a}})}^2\\ & ={\displaystyle \frac{b^2-4ac}{4a^2}}\end{array}& \begin{array}{rl}& \text{Dari bentuk 2, kita}\\ & \text{akan mendapatkan}\\ & {(x+{\displaystyle \frac{b}{2a}})}^2\\ & ={\displaystyle \frac{b^2-4ac}{4a^2}}\\ & x-{\displaystyle \frac{b}{2a}=\pm \sqrt{{\displaystyle \frac{b^2-4ac}{4a^2}}}}\\ & x=-{\displaystyle \frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}}\\ & x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & \\ & \end{array}\\ \hline\end{array}$.

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 $D=b^2-4ac$.

$\begin{array}{|ccl|}\hline \text{No}& \text{Jenis nilai}\mathbf{\text{D}}& \text{Penjelasan nilai}\mathbf{\text{D}}\\ 1& \mathbf{\text{D}}>0& \text{Persamaan kuadrat mempunyai dua akar }\\ & & \text{riil dan}\text{berbeda}\\ 2& \mathbf{\text{D}}=0& \text{Persamaan kuadrat mempunyai dua akar }\\ & & \text{riil dan}\mathit{\text{sama}}\\ 3& \mathbf{\text{D}}<0& \text{Persamaan kuadrat mempunyai dua akar }\\ & & \text{tidak riil dan}\text{berbeda}\\ \hline\end{array}$.

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

$\begin{array}{|ccl|}\hline \text{No}& \text{Kondisi akar-akar}{x}_1\mathrm{\&}{x}_2& \text{Dari posisi}a{x}^2+bx+c=0\\ 1& x_1+x_2=-{\displaystyle \frac{b}{a}}& \text{akar-akarnya tidak harus}{x}_1\mathrm{\&}{x}_2\\ & & \text{terkadang dituliskan dengan}\alpha \text{dan}\beta \\ 2& x_1\times x_2={\displaystyle \frac{c}{a}}& \text{Baik rumus jumlah maupun hasil kali}\\ & & \text{Anda juga dapat melihat dari jenis akarnya}\\ 3& x_1-x_2=\left|{\displaystyle \frac{\sqrt{D}}{a}}\right|& \text{Ingat nilai}D=b^2-4ac\\ \hline\end{array}$.

A. 4.  Menyusun Persamaan Kuadrat Baru

Persamaan kuadrat dengan akar-akar $x_{1}dan{x}_2$  dapat disusun dengan rumus:

$x^2-(x_1+x_2)x+x_1\times x_2=0$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan kar-akar dari persamaan kuadrat}\\ & (\text{a})x^2-2x-8=0\\ & (\text{b})2x^2-3x-5=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline (\mathbf{\text{a}})& (\mathbf{\text{b}})\\ \begin{array}{rl}& x^2-2x-8=0\\ & \iff (x-4)(x+2)=0\\ & \iff x-4=0\text{atau}x+2=0\\ & \iff x=4\text{atau}x=-2\\ & \\ & \\ & \end{array}& \begin{array}{rl}& 2x^2-3x-5=0\\ & \iff {\displaystyle \frac{(2x-5)(2x+2)}{2}=0}\\ & \iff (2x-5)(x+1)=0\\ & \iff 2x-5=0\text{atau}x+1=0\\ & \iff 2x=5\text{atau}x=-1\\ & \iff x={\displaystyle \frac{5}{2}\text{atau}x=-1}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Penyelesaian terkecil dari persamaan kuadrat}\\ & (x-{\displaystyle \frac{3}{4}})(x-{\displaystyle \frac{3}{4}})+(x-{\displaystyle \frac{3}{4}})(x-{\displaystyle \frac{1}{2}})=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{3}{4}})+\left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{1}{2}})=0\\ & \iff \left(x-{\displaystyle \frac{3}{4}}\right)(x-{\displaystyle \frac{3}{4}+x-\frac{1}{2}})=0\\ & \iff \left(x-{\displaystyle \frac{3}{4}}\right)(2x-{\displaystyle \frac{5}{4}})=0\\ & \iff x-{\displaystyle \frac{3}{4}=0\text{atau}2x-\frac{5}{4}=0}\\ & \iff x={\displaystyle \frac{3}{4}\text{atau}2x=\frac{5}{4}}\\ & \iff x={\displaystyle \frac{6}{8}\text{atau}x=\frac{5}{8}}\end{array}\\ & \text{Jadi, nilai terkecilnya adalah}{\displaystyle \frac{5}{8}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika persamaan}2x^2-3x-14=0\text{mempunyai}\\ & \text{akar-akar}{x}_1\text{dan}{x}_2\text{dengan}{x}_1>x_2,\text{maka}\\ & 2x_1+3x_2\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa PK}:2x^2-3x-14=0\\ & \begin{array}{rl}& 2x^2-3x-14=0\iff {\displaystyle \frac{(2x-7)(2x+4)}{2}=0}\\ & \iff (2x-7)(x+2)=0\\ & \iff 2x-7=0\text{atau}x+2=0\\ & x={\displaystyle \frac{7}{2}\text{atau}x=-2}\\ & \text{Karena nilai dari}{x}_1>x_2,\text{maka}\\ & x_1={\displaystyle \frac{7}{2}\text{dan}{x}_2=-2}\end{array}\\ & \text{Sehingga nilai}2x_1+3x_2=2{\displaystyle \frac{7}{2}+3(-2)=7-6=1}\end{array}$.

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

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

$\begin{array}{l}\\ 6.& \text{Tentukanlah akar-akar persamaan kuadrat dengan rumus kuadrat}\\ & \begin{array}{l}\\ \text{a}.x^2-2=0& \text{f}.2p^2-5p-12=0\\ \text{b}.x^2+3x-1=0& \text{g}.3q^2-11q+10=0\\ \text{c}.x^2+2x-3=0& \text{h}.4x^2+11x+6=0\\ \text{d}.x^2+5x-6=0& \text{i}.5z^2-z-4=0\\ \text{e}.x^2-7x-8=0& \text{j}.6x^2+17x+7=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.x^2& -2=0\\ & \{\begin{array}{ll}a& =1\\ b& =0\\ c& =-2\end{array}\\ x_{1,2}& ={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ x_{1,2}& ={\displaystyle \frac{0\pm \sqrt{0^2-4.1.(-2)}}{2.1}}\\ & ={\displaystyle \frac{\pm \sqrt{8}}{2}=\frac{\pm \sqrt{4.2}}{2}}\\ & ={\displaystyle \frac{\pm 2\sqrt{2}}{2}}\\ & =\pm \sqrt{2}\\ x_1& =\sqrt{2}\text{atau}{x}_2=-\sqrt{2}\end{array}& \begin{array}{rl}\text{i}.5z^2& -z-4=0\\ & \{\begin{array}{ll}a& =5\\ b& =-1\\ c& =-4\end{array}\\ z_{1,2}& ={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ z_{1,2}& ={\displaystyle \frac{-(-1)\pm \sqrt{(-1{)}^2-4.5.(-4)}}{2.5}}\\ & ={\displaystyle \frac{1\pm \sqrt{1+80}}{10}=\frac{1\pm \sqrt{81}}{10}}\\ & ={\displaystyle \frac{1\pm 9}{10}}\\ z_1& ={\displaystyle \frac{1+9}{10}=1\text{atau}{z}_2=\frac{1-9}{10}=\frac{-8}{10}=-\frac{4}{5}}\\ & \end{array}\\ \hline\end{array}\end{array}$.

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

$\begin{array}{l}\\ 8.& \text{Carilah nilai}x\text{yang memenuhi persamaan}\\ & {\displaystyle \frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}-\frac{2}{x^2-10x-69}=0}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}}& =\frac{2}{x^2-10x-69}\\ {\displaystyle \frac{1}{(x^2-10x-37)+8}+\frac{1}{(x^2-10x-37)-8}}& =\frac{2}{(x^2-10x-37)-32}\\ \text{Misalkan}{x}^2-10x-37=p,\text{maka}& \\ {\displaystyle \frac{1}{p+8}+\frac{1}{p-8}}& =\frac{2}{p-32}\\ {\displaystyle \frac{p-8+p+8}{(p+8)(p-8)}}& =\frac{2}{p-32}\\ {\displaystyle \frac{2p}{(p+8)(p-8)}}& =\frac{2}{p-32}\\ {\displaystyle \frac{p}{p^2-64}}& =\frac{1}{p-32}\\ p^2-32p& =p^2-64\\ p& ={\displaystyle \frac{-64}{-32}}\\ p& =2,\\ \text{kita kembali ke bentuk semula}& \\ x^2-10x-37& =2\\ x^2-10x-39& =0\\ (x-13)(x+3)& =0\\ x=13\text{atau}x=-3& \end{array}\\ & \text{Jadi},x=13\end{array}$.

$\begin{array}{l}\\ 9.& \text{Diketahui akar-akar persamaan kuadrat}\\ & x^2+x-3=0\text{adalah}\alpha \text{dan}\beta .\text{Tentukanlah nilai dari}\\ & {\alpha }^3-4{\beta }^2+19\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & \begin{array}{|lll|}\hline x^2+x-3=0& \begin{array}{rl}{\alpha }^2+\alpha -3& =0\\ & \\ \iff {\alpha }^2& =3-\alpha .....(1)\end{array}& \begin{array}{rl}{\beta }^2+\beta -3& =0\\ & \\ \iff {\beta }^2& =3-\beta .....(2)\end{array}\\ \{\begin{array}{c}\alpha +\beta ={\displaystyle \frac{-b}{a}=-1}\\ \alpha \beta ={\displaystyle \frac{c}{a}=-3}\end{array}& \begin{array}{rl}{\alpha }^3+{\alpha }^2-3\alpha & =0\\ & \\ \iff {\alpha }^3& =3\alpha -{\alpha }^2.....(3)\end{array}& \begin{array}{rl}{\beta }^3+{\beta }^2-3\beta & =0\\ & \\ \iff {\beta }^3& =3\beta -{\beta }^2.....(4)\end{array}\\ \hline\end{array}\\ & \\ & \begin{array}{rl}{\alpha }^3-4{\beta }^2+19& =(3\alpha -{\alpha }^2)-4(3-\beta )+19,\text{perhatikan persamaan}(3)\text{dan}(2)\\ & =3\alpha -(3-\alpha )-12+4\beta +19\\ & =4\alpha +4\beta -3+7\\ & =4(\alpha +\beta )+4\\ & =4(-1)+4\\ & =0\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{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}\\ & \text{adalah}p+\sqrt{q+\sqrt{r}},\text{dengan}p,q,\text{dan}r\\ & \text{adalah bilangan-bilangan asli}.\text{Carilah hasil}p+q+r\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\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\\ (x^2-11x)(\frac{2}{x^2-22x+57}+\frac{2}{x^2-22x+85})& =x^2-11x,\text{misal}t=x^2-22x\\ (\frac{2}{t+57}+\frac{2}{t+85})& =\frac{x^2-11x}{x^2-11x}=1\\ 2(t+85)+2(t+57)& =(t+57)(t+85)\\ 2t+170+2t+114& =t^2+142t+4845\\ 0& =t^2+138t+4731\\ t^2+138t+4731& =0\{\begin{array}{c}a=1\\ b=138\\ c=4731\end{array}\\ 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{array}\\ & \begin{array}{rl}\text{Selanjutnya}& \\ 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\text{atau}{x}^2-22x+69-\sqrt{30}\\ & \\ \text{dengan cara yang}& \text{semisal diatas}\\ & \\ x_{1,2}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69+\sqrt{30})}}{2}}& \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69-\sqrt{30})}}{2}}\\ x_{1,2}={\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}}& \text{atau}{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}}& \text{atau}{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}}& \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}}\\ x_{1,2}=11\pm \sqrt{52-\sqrt{30}}& \text{atau}{x}_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ & \\ \text{Maka},& \\ \{\begin{array}{c}{x}_1=11+\sqrt{52-\sqrt{30}}\\ \\ x_2=11-\sqrt{52-\sqrt{30}}\end{array}& \text{atau}\{\begin{array}{c}{x}_3=11+\sqrt{52+\sqrt{30}}\\ \\ x_4=11-\sqrt{52+\sqrt{30}}\end{array}\end{array}\\ \\ \\ & \begin{array}{rl}\text{Selanjutnya nilai}& \text{yang paling pas sesuai soal adalah}{x}_3=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \text{Sehingga nilai}& p+q+r=11+52+30=93\end{array}\end{array}$

Sumber Referensi:

1. Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.
2. Idris, M., Rusdi, 1. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA. 
3. Kurnianingsih, Sri, Kuntarti & Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1. Jakarta: ESIS.
4. Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: YUDISTIRA.
5. Sobirin. 2005. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 1. Jakarta: KAWAN PUSTAKA.

KUMPULAN MATERI SMA/MA KELAS X/FASE E (Bagian 2) Tahun 2024

 Kelas X 

Kurikulum Merdeka

Fase E Kelas X

Semester Gasal

Eksponen dan Logaritma

• materi eksponen 1, materi fungsi eksponen 2, lanjutan fungsi eksponen, lanjutan materi 2 fungsi ekponen, lanjutan materi 3 fungsi eksponen, Persamaan eksponen.
• contoh soal eksponen 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10 
• materi logaritma 
• materi logaritma lanjutan
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• materi logaritma lanjutan 2
• materi persamaan logaritma 1, persamaan 2, persamaan 3, persamaan 4, persamaan 5. aplikasi logaritma, pengayaan (operasi logaritma natural)
• contoh soal 7 , contoh 8 , contoh 9 , contoh 10, contoh 11, contoh 12, contoh 13

Barisan dan Deret

(pilih  materi yang Anda butuhkan saja)

• materi pola bilangan - induksi matematika
• pola bilangan dan barisan serta deret aritmetika
• pola bilangan dan barisan serta deret geometri 
• berikut contoh soal induksi matematika dan pola bilangan
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5
• materi barisan dan deret aritmetika (hitung)
• lanjutan materi barisan dan deret geometri (ukur)
• berikut contoh soalnya
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7.
• barisan dan deret aritmetika dan geometri sekaligus
• contoh soal selingan (diselipkan soal untuk kompetisi) yang terkait barisan dan deret: contoh 1, contoh 2, contoh 3, contoh 4

Trigonometri

• materi trigonometri awal
• perbandingan trigonometri pada segitiga siku-siku

• koordinat kartesius, koordinat kutub serta sudut-sudut di berbagai kuadran

• aturan sinus
• aturan cosinus (bagian 1), aturan cosinus (bagian 2)
• contoh soal

Sistem Persamaan dan Pertidaksamaan Linear

• materi persamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 
• materi pertidaksamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , 

• materi pertidaksamaan rasional dan irasional satu variabel
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , 

• materi sistem persamaan linear tiga variabel (SPLTV)
• lanjutan materi SPLTV
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

Semester Genap

• Persamaan Kuadrat dan Fungsi Kuadrat, contoh soal PK 1, contoh PK 2, contoh PK 3, contoh PH 4, contoh soal FK 1, contoh FK 2, contoh FK 3, contoh FK 4, contoh FK 5, contoh FK 6.
• Statistikaa materi statistika, penyajian data dalam bentuk diagram. ukuran pemusatan data. ukuran letak data 1, ukuran letak data 2, ukuran letak data 3, interpolasi linear, ukuran penyebaran data 1, ukuran penyebaran data 2, koefisien variansi contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10, contoh 11, contoh 12, contoh 13
• Aturan Pencacahan dan Peluang

• Aturan Pencacahan

  • materi pendahuluan(kombinatorial dan percobaan), kaidah pencacahan(kaidah perkalian), faktorial-permutasi-kombinasi, 
  • contoh soal 1 permutasi-kombinasi, contoh 2, contoh 3, 
  • materi Binomial Newton (pengayaan), fungsi pembangkit untuk kombinasi (pengayaan), 
  • contoh soal 1 (kaidah pencacahan), contoh 2, contoh 3,

  Peluang Kejadian Majmuk

  • materi pendahuluan (percobaan, ruang sampel, dan kejadian), peluang kejadian tunggal dan frekuensi harapan, peluang kejadian majmuk (peluang komplemen, dua kejadian saling bebas, dua kejadian tidak saling bebas), kejadian saling lepas, kejadian tidak saling lepas, 
  • contoh soal 1, contoh 2, contoh 3, 

Kelas XI dan Kelas XII 

Matematika Wajib

Matematika Peminatan