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

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

$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: x^{2}+2\left | x \right |-8=0\: ,\: \textrm{maka nilai}\: \: x\\ &\textrm{yang memenuhi adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -4&&&&&\textrm{D}.&\displaystyle 0\\ \textrm{B}.&\color{red}\displaystyle -2&&\textrm{C}.&\displaystyle -1&&\textrm{E}.&4 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &x^{2}+2\left | x \right |-8=0\Leftrightarrow \left ( \left | x \right |+4 \right )\left ( \left | x \right |-2 \right )=0\\ &\Leftrightarrow \left | x \right |=-4\: (\textrm{bukan solusi})\: \: \textrm{atau}\: \: \left | x \right |=2\: (\textrm{solusi})\\ &\textrm{Pilih}\: \: \left | x \right |=2\Rightarrow x=\color{red}\pm 2 \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{akar-akar dari persamaan}\\ &x^{2}-2x=\left | x-1 \right |+5,\: \textrm{maka nilai }\\ &\alpha +\beta\: \:  \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -2&&&&&\textrm{D}.&\displaystyle 1\\ \textrm{B}.&\displaystyle -1&&\textrm{C}.&\displaystyle 0&&\textrm{E}.&\color{red}2 \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &\begin{aligned}&x^{2}-2x=\left | x-1 \right |+5\Leftrightarrow x^{2}-2x-5=\left | x-1 \right |\\ &\color{blue}\textrm{Untuk}\: \: \color{black}x\geq 1,\color{blue}\: \: \textrm{persamaan akan menjadi}\\ &x^{2}-2x-5=x-1\Leftrightarrow x^{2}-2x-x-5+1=0\\ &x^{2}-3x-4=0\Leftrightarrow (x-4)(x+1)=0\\ &\Leftrightarrow x=4\: (\textrm{memenuhi})\: \: \textrm{atau}\: \: x=-1\: (\textrm{tidak})\\ &\color{blue}\textrm{Untuk}\: \: \color{black}x<1,\color{blue}\: \: \textrm{persamaan akan menjadi}\\ &x^{2}-2x-5=1-x\Leftrightarrow x^{2}-2x+x-5-1=0\\ &x^{2}-x-6=0\Leftrightarrow (x-3)(x+2)=0\\ &\Leftrightarrow x=3\: (\textrm{tidak})\: \: \textrm{atau}\: \: x=-2\: (\textrm{memenuhi})\\ &\color{blue}\textrm{Pilih}\: \: \alpha =4,\: \textrm{dan}\: \: \beta =-2,\: \textrm{maka}\\ &\alpha +\beta =4+(-2)=\color{red}2   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Persamaan kuadrat}\: \: x^{2}-2x+m=0 \: \: \textrm{mempunyai }\\ &\textrm{akar-akar yang rasional, maka nilai}\: \: m\: \: \textrm{yang}\\ &\textrm{mungkin adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}1-\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&&\\ \textrm{B}.&1+\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{C}.&\displaystyle \frac{k^{2}}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{D}.&\displaystyle \frac{k^{2}-1}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \textrm{E}.&1-\displaystyle \frac{k}{4}&\textrm{untuk}&k=0,1,2,\cdots &&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Akar-akar dari PK}:x^{2}-2x+m=0\\ &x_{1,2}=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{2\pm \sqrt{4-4m}}{2}\\  &\textrm{Agar nilai}\: \: m\: \: \textrm{rasional, maka}\\ &4-4m=k^{2}\Leftrightarrow 4m=4-k^{2}\Leftrightarrow m=\color{red}1-\displaystyle \frac{k^{2}}{4}  \end{aligned}  \end{array}$.

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