Question

Find the roots of the quadratic equation $4x^2+4\sqrt{3}x+3=0$ by using quadratic formula.

• A. $-\frac{3}{2},\frac{3}{2}$
• B. $-\frac{3}{2},-\frac{3}{2}$
• C. $-\frac{\sqrt{3}}{2}$, $+\frac{\sqrt{3}}{2}$
• D. $-\frac{\sqrt{3}}{2}$, $-\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D $-\frac{\sqrt{3}}{2}$, $-\frac{\sqrt{3}}{2}$

$4x^2+4\sqrt{3}x+3=0$

On comparing this equation with $a{x}^2+bx+c=0$, we get
$a=4,b=4\sqrt{3}andc=3$

By using quadratic formula, we get

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

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

$\Rightarrow$ $x=\frac{-4\sqrt{3}\pm \sqrt{48-48}}{8}$

$\Rightarrow$ $x=\frac{-4\sqrt{3}\pm \sqrt{0}}{8}$

$\Rightarrow$ $x=\frac{-4\sqrt{3}}{8}$

$\therefore x=-\frac{\sqrt{3}}{2}$

We see that the discriminant is 0.
So, the roots of the equation are equal.

Hence the roots of $4x^2+4\sqrt{3}x+3=0$ are $-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$.