Question

Solve the quadratic: $\sqrt{2}{x}^2+x+\sqrt{2}=0$

Answer 1

We have, $\sqrt{2}{x}^2+x+\sqrt{2}=0$
On comparing with general form of quadratic equation:

$a{x}^2+bx+c=0$

We get $a=\sqrt{2},b=-1,c=\sqrt{2}$

Substituting these values in

$\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}and\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$

We will obtain

$\alpha =\frac{-1+\sqrt{1^2-4\times \sqrt{2}\times \sqrt{2}}}{2\times \sqrt{2}}$

$=\frac{-1+\sqrt{1-8}}{2\times \sqrt{2}}=\frac{-1+\sqrt{-7}}{2\sqrt{2}}$

$\alpha =\frac{-1+\sqrt{(-1)7}}{2\sqrt{2}}=\frac{-1+\sqrt{7}i}{2\sqrt{2}}[\because i=\sqrt{(-1)}]$

$and\beta =\frac{-1-\sqrt{1^2-4\times \sqrt{2}\times \sqrt{2}}}{2\times \sqrt{2}}=\frac{-1-\sqrt{7}i}{2\sqrt{2}}$

Hence, the roots are $\frac{-1+\sqrt{7}i}{2\sqrt{2}}\&\frac{-1-\sqrt{7}i}{2\sqrt{2}}$