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In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
$3x^2+2\sqrt5 x-5=0$
Given:
Given quadratic equation is $3x^2+2\sqrt5 x-5=0$.
To do:
We have to determine whether the given quadratic equation has real roots.
Solution:
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=3, b=2\sqrt5$ and $c=-5$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is
$D=b^2-4ac$.
Therefore,
$D=(2\sqrt5)^2-4(3)(-5)=4(5)+60=20+60=80$.
As $D>0$, the given quadratic equation has real roots and the roots are
$x=\frac{-b\pm \sqrt{D}}{2a}$
$x=\frac{-2\sqrt5\pm \sqrt{80}}{2(3)}$
$x=\frac{-2\sqrt5\pm \sqrt{4\times4\times5}}{2(3)}$
$x=\frac{-2\sqrt5\pm 4\sqrt5}{6}$
$x=\frac{-2(\sqrt5\pm 2\sqrt5)}{2\times3}$
$x=\frac{-(\sqrt5\pm 2\sqrt5)}{3}$
$x=\frac{-(\sqrt5+2\sqrt5)}{3}$ or $x=\frac{-(\sqrt5-2\sqrt5)}{3}$
$x=\frac{-\sqrt5-2\sqrt5}{3}$ or $x=\frac{-\sqrt5+2\sqrt5}{3}$
$x=\frac{-3\sqrt5}{3}$ or $x=\frac{\sqrt5}{3}$
$x=\frac{\sqrt5}{3}$ or $x=-\sqrt5$
The roots are $\frac{\sqrt5}{3}$ and $-\sqrt5$.