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Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: 36 x^{2}-12 a x+\left(a^{2}-b^{2}\right)=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: 36 x^{2}-12 a x+\left(a^{2}-b^{2}\right)=0

The given equation is 36 x^{2}-12 a x+\left(a^{2}-b^{2}\right)=0

Comparing it with A x^{2}+B x+C=0, we get
A=36, B=-12 a and C=a^{2}-b^{2}

\therefore Discriminant,

D=B^{2}-4 A C=(-12 a)^{2}-4 \times 36 \times\left(a^{2}-b^{2}\right)=144 a^{2}-144 a^{2}+144 b^{2}=144 b^{2}>0

So, the given equation has real roots

Now, \sqrt{D}=\sqrt{144 b^{2}}=12 b

\begin{array}{l} \therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-(-12 a)+12 b}{2 \times 36}=\frac{12(a+b)}{72}=\frac{a+b}{0} \\ \beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-(-12 a)-12 b}{2 \times 36}=\frac{12(a-b)}{72}=\frac{a-b}{6} \end{array}

Hence, \frac{a+b}{6} and \frac{a-b}{6} are the roots of the given equation.

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