Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:$3x^2 - 4\sqrt3x + 4 = 0$

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Given:

Given quadratic equation is $3x^2 - 4\sqrt3x + 4 = 0$.

To do:

We have to find the nature of the roots of the given quadratic equation and find them.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=3, b=-4\sqrt3$ and $c=4$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

Therefore,

$D=(-4\sqrt3)^2-4(3)(4)=16(3)-12(4)$

$=48-48$

$=0$

As $D=0$, the given quadratic equation has real and equal roots.

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

$=\frac{-(-4 \sqrt{3}) \pm 0}{2 \times 3}$

$=\frac{4 \sqrt{3}}{6}$

Hence, the roots of the given quadratic equation are $\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}$

Updated on 10-Oct-2022 13:20:13