Question

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

• A. $1,i$
• B. $\frac{(1-\surd 7i)}{2},0$
• C. $1,-i$
• D. $\frac{(-1+\sqrt{7}i)}{2\sqrt{2}},\frac{(-1-\sqrt{7}i)}{2\sqrt{2}}$

Answer 1

Answer: (D)

$\frac{(-1+\sqrt{7}i)}{2\sqrt{2}},\frac{(-1-\sqrt{7}i)}{2\sqrt{2}}$

We have, $x^2+\frac{x}{\sqrt{2}}+1=0$

$\Rightarrow \sqrt{2}{x}^2+x+\sqrt{2}=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1-4\left(\sqrt{2}\right)\left(\sqrt{2}\right)}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-1\pm \sqrt{1-8}}{2\sqrt{2}}$

$\Rightarrow x=\frac{-1\pm 7i}{2\sqrt{2}}$