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