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

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

$\begin{array}{l}\\ 12.& \text{Jika}{x}^2+2\left|x\right|-8=0,\text{maka nilai}x\\ & \text{yang memenuhi adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -4}& & & & & \text{D}.& {\displaystyle 0}\\ \text{B}.& {\displaystyle -2}& & \text{C}.& {\displaystyle -1}& & \text{E}.& 4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & x^2+2\left|x\right|-8=0\iff (\left|x\right|+4)(\left|x\right|-2)=0\\ & \iff \left|x\right|=-4(\text{bukan solusi})\text{atau}\left|x\right|=2(\text{solusi})\\ & \text{Pilih}\left|x\right|=2\Rightarrow x=\pm 2\end{array}$.

$\begin{array}{l}\\ 13.& \text{Jika}\alpha \text{dan}\beta \text{akar-akar dari persamaan}\\ & x^2-2x=|x-1|+5,\text{maka nilai }\\ & \alpha +\beta \text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -2}& & & & & \text{D}.& {\displaystyle 1}\\ \text{B}.& {\displaystyle -1}& & \text{C}.& {\displaystyle 0}& & \text{E}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & \begin{array}{rl}& x^2-2x=|x-1|+5\iff x^2-2x-5=|x-1|\\ & \text{Untuk}x\ge 1,\text{persamaan akan menjadi}\\ & x^2-2x-5=x-1\iff x^2-2x-x-5+1=0\\ & x^2-3x-4=0\iff (x-4)(x+1)=0\\ & \iff x=4(\text{memenuhi})\text{atau}x=-1(\text{tidak})\\ & \text{Untuk}x<1,\text{persamaan akan menjadi}\\ & x^2-2x-5=1-x\iff x^2-2x+x-5-1=0\\ & x^2-x-6=0\iff (x-3)(x+2)=0\\ & \iff x=3(\text{tidak})\text{atau}x=-2(\text{memenuhi})\\ & \text{Pilih}\alpha =4,\text{dan}\beta =-2,\text{maka}\\ & \alpha +\beta =4+(-2)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Persamaan kuadrat}{x}^2-2x+m=0\text{mempunyai }\\ & \text{akar-akar yang rasional, maka nilai}m\text{yang}\\ & \text{mungkin adalah}....\\ & \begin{array}{lllllll}\text{A}.& 1-{\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & & \\ \text{B}.& 1+{\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{C}.& {\displaystyle \frac{k^2}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{D}.& {\displaystyle \frac{k^2-1}{4}}& \text{untuk}& k=0,1,2,\cdots & & \\ \text{E}.& 1-{\displaystyle \frac{k}{4}}& \text{untuk}& k=0,1,2,\cdots & & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Akar-akar dari PK}:x^2-2x+m=0\\ & x_{1,2}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{2\pm \sqrt{4-4m}}{2}}\\ & \text{Agar nilai}m\text{rasional, maka}\\ & 4-4m=k^2\iff 4m=4-k^2\iff m=1-{\displaystyle \frac{k^2}{4}}\end{array}\end{array}$.

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