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.