Question

Solve each of the following quadratic equations:

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

Answer 1

We write, $4bx=2\left(a+b\right)x-2\left(a-b\right)x$ as $4x^2\times \left[-\left(a^2-b^2\right)\right]=-4\left(a^2-b^2\right)x^2=2\left(a+b\right)x\times \left[-2\left(a-b\right)x\right]$
$$\begin{gathered} \therefore 4x^2+4bx-\left(a^2-b^2\right)=0 \\ \Rightarrow 4x^2+2\left(a+b\right)x-2\left(a-b\right)x-\left(a-b\right)\left(a+b\right)=0 \\ \Rightarrow 2x\left[2x+\left(a+b\right)\right]-\left(a-b\right)\left[2x+\left(a+b\right)\right]=0 \\ \Rightarrow \left[2x+\left(a+b\right)\right]\left[2x-\left(a-b\right)\right]=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x+\left(a+b\right)=0\mathrm{or}2x-\left(a-b\right)=0 \\ \Rightarrow x=-\frac{a+b}{2}\mathrm{or}x=\frac{a-b}{2} \end{gathered}$$
Hence, $-\frac{a+b}{2}$ and $\frac{a-b}{2}$ are the roots of the given equation.