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Find the roots of the 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 | Class 10 | Exercise 10C | Maths | Quadratic Equations | RS Aggarwal
Find the roots of the 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.

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}

\therefore 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}

\begin{array}{l} \therefore \alpha=\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} \\ \beta=\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} \end{array}

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

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