$\begin{array}{ll}\\ 1.&\textrm{Penyelesaian terkecil dari persamaan kuadrat}\\ & \left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( x-\displaystyle \frac{3}{4} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&-\displaystyle \frac{3}{4}&&&&&\textrm{D}.&\displaystyle \frac{3}{4}\\ \textrm{B}.&\displaystyle \frac{1}{2}&&\textrm{C}.&\color{red}\displaystyle \frac{5}{8}&&\textrm{E}.&1 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4} \right )+\left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( x-\displaystyle \frac{3}{4}+x-\frac{1}{2} \right )=0\\ &\Leftrightarrow \left ( \color{blue}x-\displaystyle \frac{3}{4}\color{black} \right )\left ( 2x-\displaystyle \frac{5}{4} \right )=0\\ &\Leftrightarrow x-\displaystyle \frac{3}{4}=0\: \: \textrm{atau}\: \: 2x-\frac{5}{4}=0\\ &\Leftrightarrow x=\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 2x=\frac{5}{4}\\ &\Leftrightarrow x=\displaystyle \frac{6}{8}\: \: \textrm{atau}\: \: x=\frac{5}{8}\\ \end{aligned} \\ &\textrm{Jadi, nilai terkecilnya adalah}\: \: \color{red}\displaystyle \frac{5}{8} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: m\: \: \textrm{dan}\: \: n\: \: \textrm{adalah penyelesaian dari}\\ &\textrm{persamaan kuadrat}\: \: x^{2}+mx+n=0,\\ &\textrm{dengan}\: \: m\neq 0\: \: \textrm{dan}\: \: n\neq 0\: \: \textrm{jumlah kedua}\\ &\textrm{penyelesaian tersebut adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -\frac{1}{2}&&&&&\textrm{D}.&\displaystyle 1\\ \textrm{B}.&\color{red}\displaystyle -1&&\textrm{C}.&\displaystyle \frac{1}{2}&&\textrm{E}.&\displaystyle \textrm{tidak dapat ditentukan} \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa PK}:x^{2}+mx+n=0\\ &\textrm{dengan}\: \: a=1,\: b=m,\: \: \textrm{dan}\: \: c=n\\ &\begin{aligned}&\bullet \quad x_{1}+ x_{2}=-\displaystyle \frac{b}{a}\Leftrightarrow m+n=-\frac{m}{1}\\ &\bullet \quad x_{1}\times x_{2}=\displaystyle \frac{c}{a}\quad\Leftrightarrow mn=\frac{n}{1}\Leftrightarrow m=1\\ &\textbf{Dari persamaan pertama akan diperoleh}\\ &m+n=-m=\color{red}-1 \end{aligned}\\ &\textrm{Jadi, nilai m+n adalah}\: \: \color{red}\displaystyle -1 \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Jika persamaan}\: \: 2x^{2}-3x-14=0\: \: \textrm{mempunyai}\\&\textrm{akar-akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{dengan}\: \: x_{1}>x_{2}, \: \: \textrm{maka}\\ &2x_{1}+3x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 2&&&&&\textrm{D}.&\displaystyle -2\\ \textrm{B}.&\color{red}\displaystyle 1&&\textrm{C}.&\displaystyle -1&&\textrm{E}.&-5 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa PK}\: :\: 2x^{2}-3x-14=0\\ &\begin{aligned}&2x^{2}-3x-14=0\Leftrightarrow \displaystyle \frac{(2x-7)(2x+4)}{2}=0\\ &\Leftrightarrow (2x-7)(x+2)=0\\ &\Leftrightarrow 2x-7=0\: \: \textrm{atau}\: \: x+2=0\\ &x=\displaystyle \frac{7}{2}\: \: \textrm{atau}\: \: x=-2\\ &\textrm{Karena nilai dari}\: \: x_{1}>x_{2},\: \: \textrm{maka}\\ &x_{1}=\displaystyle \frac{7}{2}\: \: \textrm{dan}\: \: x_{2}=-2\\ \end{aligned}\\ &\textrm{Sehingga nilai}\: \: 2x_{1}+3x_{2}=\color{blue}2\displaystyle \frac{7}{2}+3(-2)\color{black}=7-6=\color{red}1 \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{adalah akar-akar dari}\\ &\textrm{persamaan kuadrat}\: \: x^{2}+6x+2=0,\\ &\textrm{nilai dari}\: \: x_{1}^{2}+x_{2}^{2}-4x_{1}x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 28&&&&&\textrm{D}.&\displaystyle 18\\ \textrm{B}.&\displaystyle 26&&\textrm{C}.&\color{red}\displaystyle 24&&\textrm{E}.&\displaystyle 16 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\textrm{Diketahui bahwa PK}:x^{2}+6x+2=0\\ &\textrm{dengan}\: \: a=1,\: b=6,\: \: \textrm{dan}\: \: c=2\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\color{blue}x_{1}^{2}+x_{2}^{2}\color{black}-4x_{1}x_{2}\\ &=\color{blue}\left ( x_{1}+x_{2} \right )^{2}-2x_{1}x_{2}\color{black}-4x_{1}x_{2}\\ &=\left ( x_{1}+x_{2} \right )^{2}-6x_{1}x_{2}\\ &=\left ( -\displaystyle \frac{b}{a} \right )^{2}-6\left ( \displaystyle \frac{c}{a} \right )=(-6)^{2}-6(2)\\ &=36-12\\ &=\color{red}24 \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&x^{2}+6x+2=0\Leftrightarrow x^{2}=-6x-2\\ &\bullet \quad x=x_{1}\Rightarrow x_{1}^{2}=-6x_{1}-2\\ &\bullet \quad x=x_{2}\Rightarrow x_{2}^{2}=-6x_{2}-2\\ &\qquad \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\quad+ \\ & \Leftrightarrow \quad x_{1}^{2}+x_{2}^{2}=-6\left (x_{1}+x_{2} \right )-4\\ & \Leftrightarrow \quad x_{1}^{2}+x_{2}^{2}-4x_{1}x_{2}=-6\left (x_{1}+x_{2} \right )-4x_{1}x_{2}-4\\ &\qquad\qquad\qquad=-6\left ( -\displaystyle \frac{b}{a} \right )-4\left ( \frac{c}{a} \right )-4\\ &\qquad\qquad\qquad=-6(-6)-4(2)-4\\ &\qquad\qquad\qquad=36-8-4\\ &\qquad\qquad\qquad=\color{red}24\\ \end{aligned}\\ &\textrm{Jadi, nilai yang dimaksud adalah}\: \: \color{red}\displaystyle 24 \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Diketahui akar-akar dari persamaan}\: \: 7x=4x^{2}+3\\ &\textrm{adalah}\: \: \alpha \: \: \textrm{dan}\: \: \beta .\: \textrm{Nilai} \: \: \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \frac{25}{12}&&&&&\textrm{D}.&\displaystyle \frac{16}{25}\\\\ \textrm{B}.&\displaystyle \frac{24}{12}&&\textrm{C}.&\displaystyle \frac{20}{25}&&\textrm{E}.&\displaystyle \frac{12}{25} \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\textrm{Diketahui bahwa PK}:4x^{2}-7x+3=0\\ &\textrm{dengan}\: \: a=4,\: b=-7,\: \: \textrm{dan}\: \: c=3\\ &\begin{aligned}&\displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\displaystyle \frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }=\frac{(\alpha +\beta )^{2}-2\alpha \beta }{\alpha \beta }\\ &=\displaystyle \frac{\left (\displaystyle -\frac{b}{a} \right )^{2} -2\left (\displaystyle \frac{c}{a} \right )}{\displaystyle \frac{c}{a}}=\displaystyle \frac{\left ( \displaystyle \frac{7}{4} \right )^{2}-2\left ( \displaystyle \frac{3}{4} \right )}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{\displaystyle \frac{49}{16}-\frac{6}{4}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{\displaystyle \frac{49-24}{16}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{\displaystyle \frac{25}{16}}{\displaystyle \frac{3}{4}}=\displaystyle \frac{25}{16}\times \frac{4}{3}\\ &=\color{red}\displaystyle \frac{25}{12} \end{aligned} \end{array}$.
- Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.
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