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.

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.