Question:
Solve the following quadratic equations by factorization:
$x^{2}-4 \sqrt{2} x+6=0$
Solution:
We have been given
$x^{2}-4 \sqrt{2} x+6=0$
$x^{2}-3 \sqrt{2} x-\sqrt{2} x+6=0$
$x(x-3 \sqrt{2})-\sqrt{2}(x-3 \sqrt{2})=0$
$(x-\sqrt{2})(x-3 \sqrt{2})=0$
Therefore,
$x-\sqrt{2}=0$
$x=\sqrt{2}$
or,
$x-3 \sqrt{2}=0$
$x=3 \sqrt{2}$
Hence, $x=\sqrt{2}$ or $x=3 \sqrt{2}$.
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