Question

Solve the following quadratic equation by $\sqrt{3}{x}^2+11x+6\sqrt{3}=0$

Answer 1

Given that:

$\sqrt{3}{x}^2+11x+6\sqrt{3}=\mathrm{0........}\left(1\right)$

Since we know that if given qyadratic equation is $a{x}^2+bx+c=\mathrm{0.....}\left(2\right)$

than root are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now comparing equation 1 and 2, than

$a=\sqrt{3},b=11,c=6\sqrt{3}$

Now roots are

$x=\frac{-11\pm \sqrt{(11{)}^2-4\times (\sqrt{3})\times (6\sqrt{3})}}{2\sqrt{3}}$

$x=\frac{-11\pm \sqrt{121-72}}{2\sqrt{3}}$

$x=\frac{-11\pm \sqrt{4}9}{2\sqrt{3}}$

$x=\frac{-11\pm 7}{2\sqrt{3}}$

So Roots are,

$x_1=\frac{-11+7}{2\sqrt{3}},x_2=\frac{-11-7}{2\sqrt{3}}$

$x_1=\frac{-2}{\sqrt{3}},x_2=\frac{-9}{\sqrt{3}}$

$x_1=\frac{-2}{\sqrt{3}},x_2=-3\sqrt{3}$