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

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ [CBSE 2015]

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Solution

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

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

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

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 8 + 24 = 32 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{32} = 4 \sqrt{2}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{6 \sqrt{2}}{2 \sqrt{3}} = \sqrt{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{- 2 \sqrt{2}}{2 \sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

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