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Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: 3 a^{2} x^{2}+8 a b x+4 b^{2}=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: 3 a^{2} x^{2}+8 a b x+4 b^{2}=0

3 a^{2} x^{2}+8 a b x+4 b^{2}=0

On comparing it with A x^{2}+B x+C=0, we get:

A=3 a^{2}, B=8 a b and C=4 b^{2}

Discriminant D is given by:

\begin{array}{l} D=\left(B^{2}-4 A C\right) \\ =(8 a b)^{2}-4 \times 3 a^{2} \times 4 b^{2} \\ =16 a^{2} b^{2}>0 \end{array}

Hence, the roots of the equation are real.

Roots \alpha and \beta are given by:

\begin{array}{l} \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-8 a b+\sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}}=\frac{-8 a b+4 a b}{6 a^{2}}=\frac{-4 a b}{6 a^{2}}=\frac{-2 b}{3 a} \\ \beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-8 a b-\sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}}=\frac{-8 a b-4 a b}{6 a^{2}}=\frac{-12 a b}{6 a^{2}}=\frac{-2 b}{a} \end{array}

Thus, the roots of the equation are \frac{-2 b}{3 a} and \frac{-2 b}{a}.

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