Question
Solve the following quadratic equations by factorization:
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
✓
Answer
We have been given
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
$3\text{x}^2-\sqrt{6}\text{x}-\sqrt{6}\text{x}+2=0$
$\sqrt{3}\text{x}\big(\sqrt{3}\text{x}-\sqrt{2}\big)-\sqrt{2}\big(\sqrt{3}\text{x}-\sqrt{2}\big)=0$
$\big(\sqrt{3}\text{x}-\sqrt{2}\big)(\sqrt{3}\text{x}-\sqrt{2})=0$
Therefore,
$\sqrt{3}\text{x}-\sqrt{2}=0$
$\sqrt{3}\text{x}=\sqrt{2}$
$\text{x}=\sqrt{\frac{2}{3}}$
Hence, $\text{x}=\sqrt{\frac{2}{3}}$
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
$3\text{x}^2-\sqrt{6}\text{x}-\sqrt{6}\text{x}+2=0$
$\sqrt{3}\text{x}\big(\sqrt{3}\text{x}-\sqrt{2}\big)-\sqrt{2}\big(\sqrt{3}\text{x}-\sqrt{2}\big)=0$
$\big(\sqrt{3}\text{x}-\sqrt{2}\big)(\sqrt{3}\text{x}-\sqrt{2})=0$
Therefore,
$\sqrt{3}\text{x}-\sqrt{2}=0$
$\sqrt{3}\text{x}=\sqrt{2}$
$\text{x}=\sqrt{\frac{2}{3}}$
Hence, $\text{x}=\sqrt{\frac{2}{3}}$
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