In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

$\sqrt3 x^2+10x-8\sqrt3=0$

Given:

Given quadratic equation is $\sqrt3 x^2+10x-8\sqrt3=0$.

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,

$a=\sqrt3, b=10$ and $c=-8\sqrt3$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(10)^2-4(\sqrt3)(-8\sqrt3)=100+32(3)=100+96=196$.

As $D>0$, the given quadratic equation has real roots and the roots are

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-10\pm \sqrt{196}}{2(\sqrt3)}$ 

$x=\frac{-10\pm 14}{2\sqrt3}$ 

$x=\frac{2(-5\pm 7)}{2\sqrt3}$ 

$x=\frac{-5+7}{\sqrt3}$ or $x=\frac{-5-7}{\sqrt3}$

$x=\frac{2}{\sqrt3}$ or $x=\frac{-12}{\sqrt3}$

$x=\frac{2}{\sqrt3}$ or $x=-\frac{4\sqrt3 \times \sqrt3}{\sqrt3}$

$x=\frac{2}{\sqrt3}$ or $x=-4\sqrt3$

The roots are $\frac{2}{\sqrt3}$ and $-4\sqrt3$. 

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Updated on: 10-Oct-2022

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