Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$3 x^{2} - 2 \sqrt{6} x + 2 = 0$ [CBSE 2012]

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Solution

The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = $- 2 \sqrt{6}$ and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) + 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) - 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

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