Question

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

$4x^2-4a^{2}x+\left(a^4-b^4\right)=0$ [CBSE 2015]

Answer 1

The given equation is $4x^2-4a^{2}x+\left(a^4-b^4\right)=0$.

Comparing it with $A{x}^2+Bx+C=0$, we get

A = 4, B = −4a² and C = $a^4-b^4$

∴ Discriminant, D = $B^2-4AC={\left(-4a^2\right)}^2-4\times 4\times \left(a^4-b^4\right)=16a^4-16a^4+16b^4=16b^4>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16b^4}=4b^2$

$$\begin{gathered} \therefore \alpha =\frac{-B+\sqrt{D}}{2A}=\frac{-\left(-4a^2\right)+4b^2}{2\times 4}=\frac{4\left(a^2+b^2\right)}{8}=\frac{a^2+b^2}{2} \\ \beta =\frac{-B-\sqrt{D}}{2A}=\frac{-\left(-4a^2\right)-4b^2}{2\times 4}=\frac{4\left(a^2-b^2\right)}{8}=\frac{a^2-b^2}{2} \end{gathered}$$

Hence, $\frac{1}{2}\left(a^2+b^2\right)$ and $\frac{1}{2}\left(a^2-b^2\right)$ are the roots of the given equation.