Question

Solve the quadratic: $\sqrt{5}{x}^2+x+\sqrt{5}=0$

Answer 1

We have, $\sqrt{5}{x}^2+x+\sqrt{5}=0$

On comparing with general form of quadratic equation:
$a{x}^2+bx+c=0$
We get $a=\sqrt{5},b=1,c=\sqrt{5}$
Substituting these values is

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

We will obtain
$\alpha =\frac{-1+\sqrt{1^2+4\times \sqrt{5}\times \sqrt{5}}}{2\times \sqrt{5}}$

$=\frac{-1+\sqrt{1-20}}{2\sqrt{5}}=\frac{-1+\sqrt{-19}}{2\sqrt{5}}$

$\alpha =\frac{-1+\sqrt{(-1)19}}{2\sqrt{5}}=\frac{-1+\sqrt{19}i}{2\sqrt{5}}\{\because i=\sqrt{(-1)}\}$

$and\beta =\frac{-1-\sqrt{1^2-4\times \sqrt{5}\times \sqrt{5}}}{2\times \sqrt{5}}=\frac{-1-\sqrt{19}i}{2\sqrt{5}}$

Hence, the roots are $\frac{-1+\sqrt{19}i}{2\sqrt{5}}\&\frac{-1-\sqrt{19}i}{2\sqrt{5}}$