PAS MATEMATIKA BLOG

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

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

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

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

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

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

Sumber Referensi:

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

Fungsi Kuadrat (Kelas X/Fase E Semester 2)

B. Fungsi Kuadrat

B. 1 Fungsi

Silahkan lihat materi sebelumnya, cari di blog ini

di sini 1, dan di sini 2

B. 2 Fungsi Kuadrat

Perhatikan tabel berikut

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

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

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

B. 3 Masalah yang Melibatkan Fungsi Kuadrat

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

B.3.1 Titik Stasioner

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

B.3.2 Definit Positif dan Definit Negatif

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

Perhatikan tambahan penjelasan berikut

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

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

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

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

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

Sumber Referensi:

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

Persamaan Kuadrat (Kelas X/Fase E Semester 2)

Semester Genap

Persamaan dan Fungsi Kuadrat
Statistika
Aturan Pencacahan dan Peluang

A. Persamaan dan Fungsi Kuadrat

A. 1 Bentuk umum persamaan kuadrat

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

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

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

A. 2. Jenis-Jenis Akar Persamaan Kuadrat

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

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

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

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

A. 4. Menyusun Persamaan Kuadrat Baru

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

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

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

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

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

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

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

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

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

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

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

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

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

Sumber Referensi:

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