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

Answer

$\text{x}^2+\text{x}+\frac{1}{\sqrt{2}}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac =1^2-4.1.\frac{1}{\sqrt{2}}$$= 1 - 2\sqrt{2}$
From (A) $\text{x}=\frac{-1\pm\sqrt{-(2\sqrt{2}-1})}{2}$ $=\frac{-1\pm\sqrt{2\sqrt{2}-1\text{ i}}}{2}$ Thus, $\therefore\text{x}=\frac{-1\pm\sqrt{2\sqrt{2}-1\text{ i}}}{2}$

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