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

$2x^2-2\sqrt6 x+3=0$

Given:

Given quadratic equation is $2x^2-2\sqrt6 x+3=0$.

To do:

We have to determine whether the given quadratic equation has real roots.

Solution:

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

$a=2, b=-2\sqrt6$ and $c=3$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(-2\sqrt6)^2-4(2)(3)=4(6)-24=24-24=0$.

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

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

$x=\frac{-(-2\sqrt6)\pm \sqrt{0}}{2(2)}$ 

$x=\frac{2\sqrt6}{2(2)}$ 

$x=\frac{\sqrt2 \times \sqrt3}{\sqrt2 \times \sqrt2}$ 

$x=\frac{\sqrt3}{\sqrt2}$

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

The roots are $\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$. 

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Updated on: 10-Oct-2022

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