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