Question

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
$3x^2+2\sqrt{5}x-5=0$

Answer 1

The given equation is $3x^2+2\sqrt{5}x-5=0$

Here, $a=3,b=2\sqrt{5},c=-5$

$\therefore D=b^2-4ac={\left(2\sqrt{5}\right)}^2-4\times 3\times -5=80>0$

So, the given equation has real roots, given by

$\alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-2\sqrt{5}+\sqrt{80}}{2\times 3}=\frac{\sqrt{5}}{2}$
$\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-2\sqrt{5}-\sqrt{80}}{2\times 2}=-\sqrt{5}$