Question
Show the following quadratic equation by factorization method:
$\text{x}^2+\frac{\text{x}}{\sqrt{2}}+1=0$

Answer

$\text{x}^2+\frac{\text{x}}{\sqrt{2}}+1=0$
$\Rightarrow\sqrt{2\text{x}}^2+\text{x}+\sqrt{2}=0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
$D = b^2 - 4ac$
$=1^2-4.\sqrt{2}.\sqrt{2}$
$= 1 - 8$
$= -7$
From (A)
$\text{x}=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}$
$=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}$
Thus,
$\therefore\text{x}=\frac{-1\pm\sqrt{7}\text{ i}}{2\sqrt{2}}$

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