4-1.Complex numbers
medium

Solve the equation $x^{2}+x+\frac{1}{\sqrt{2}}=0$

A

$\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$

B

$\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$

C

$\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$

D

$\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$

Solution

The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$

This equation can also be written as $\sqrt{2} x^{2}+\sqrt{2} x+1=0$

On comparing the given equation with $a x^{2}+b x+c=0,$

we obtain $a=\sqrt{2}, b=\sqrt{2},$ and $c=1$

$\therefore $ Discriminant $(D)=b^{2}-4 a c=(\sqrt{2})^{2}-4 \times(\sqrt{2}) \times 1=2-4 \sqrt{2}$

Therefore, the required solutions are

$\frac{-b \pm \sqrt{D}}{2 a}=\frac{-\sqrt{2} \pm \sqrt{2-4 \sqrt{2}}}{2 \times \sqrt{2}}=\frac{-\sqrt{2} \pm \sqrt{2(1-2 \sqrt{2})}}{2 \sqrt{2}}$

$=\left(\frac{-\sqrt{2} \pm \sqrt{2}(\sqrt{2 \sqrt{2}-1}) i}{2 \sqrt{2}}\right) \quad[\sqrt{-1}=i]$

$=\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$

Standard 11
Mathematics

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