Question

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

${\displaystyle \sqrt{3}{x}^2+10x-8\sqrt{3}=0}$

Answer 1

We have been given, ${\displaystyle \sqrt{3}{x}^2+10x-8\sqrt{3}=0}$

Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:

D = b² - 4ac

Now, according to the equation given to us, we have,${\displaystyle a=\sqrt{3}}$, b = 10 and ${\displaystyle c=-8\sqrt{3}}$.

Therefore, the discriminant is given as,

${\displaystyle D={\left(10\right)}^2-4\left(\sqrt{3}\right)(-8\sqrt{3})}$

= 100 + 96

= 196

Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

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

Therefore, the roots of the equation are given as follows,

${\displaystyle x=\frac{-\left(10\right)\pm \sqrt{196}}{2\left(\sqrt{3}\right)}}$

${\displaystyle =\frac{-10\pm 14}{2\sqrt{3}}}$

${\displaystyle =\frac{-5\pm 7}{\sqrt{3}}}$

Now we solve both cases for the two values of x. So, we have,

${\displaystyle x=\frac{-5+7}{\sqrt{3}}}$

${\displaystyle =\frac{2}{\sqrt{3}}}$

Also,

${\displaystyle x=\frac{-5-7}{\sqrt{3}}}$

${\displaystyle =-4\sqrt{3}}$

Therefore, the roots of the equation are ${\displaystyle \frac{2}{\sqrt{3}}}$ and ${\displaystyle -4\sqrt{3}}$.