Determine the nature of the roots of the following quadratic equations:

$3x^2 - 2\sqrt6x + 2 = 0$

Given:

Given quadratic equation is $3x^2 - 2\sqrt6x + 2 = 0$.

To do:

We have to determine the nature of the roots of the given quadratic equation.

Solution:

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

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

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

Therefore,

$D=(-2\sqrt6)^2-4(3)(2)=4(6)-12(2)$

$=24-24$

$=0$

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

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Updated on: 10-Oct-2022

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