Question
Solve the following quadratic equations by factorization:
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
✓
Answer
We have been given
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}^2-4\text{x}+\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}\big(\text{x}-2\sqrt{2}\big)+1\big(\text{x}-2\sqrt{2}\big)=0$
$\big(\text{x}-2\sqrt{2}\big)\big(\sqrt{2}\text{x}+1\big)=0$
Therefore,
$\text{x}-2\sqrt{2}=0$
$\text{x}=2\sqrt{2}$
or, $\sqrt{2}\text{x}+1=0$
$\sqrt{2}\text{x}=-1$
$\text{x}=\frac{-1}{\sqrt{2}}$
Hence, $\text{x}=2\sqrt{2}$ or $\text{x}=\frac{-1}{\sqrt{2}}$
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}^2-4\text{x}+\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}\big(\text{x}-2\sqrt{2}\big)+1\big(\text{x}-2\sqrt{2}\big)=0$
$\big(\text{x}-2\sqrt{2}\big)\big(\sqrt{2}\text{x}+1\big)=0$
Therefore,
$\text{x}-2\sqrt{2}=0$
$\text{x}=2\sqrt{2}$
or, $\sqrt{2}\text{x}+1=0$
$\sqrt{2}\text{x}=-1$
$\text{x}=\frac{-1}{\sqrt{2}}$
Hence, $\text{x}=2\sqrt{2}$ or $\text{x}=\frac{-1}{\sqrt{2}}$
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