Question

Solve each of the following quadratic equations:

$x^2-\left(2b-1\right)x+\left(b^2-b-20\right)=0$ [CBSE 2015]

Answer 1

We write, $-\left(2b-1\right)x=-\left(b-5\right)x-\left(b+4\right)x$ as $x^2\times \left(b^2-b-20\right)=\left(b^2-b-20\right)x^2=\left[-\left(b-5\right)x\right]\times \left[-\left(b+4\right)x\right]$
$$\begin{gathered} \therefore x^2-\left(2b-1\right)x+\left(b^2-b-20\right)=0 \\ \Rightarrow x^2-\left(b-5\right)x-\left(b+4\right)x+\left(b-5\right)\left(b+4\right)=0 \\ \Rightarrow x\left[x-\left(b-5\right)\right]-\left(b+4\right)\left[x-\left(b-5\right)\right]=0 \\ \Rightarrow \left[x-\left(b-5\right)\right]\left[x-\left(b+4\right)\right]=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-\left(b-5\right)=0\mathrm{or}x-\left(b+4\right)=0 \\ \Rightarrow x=b-5\mathrm{or}x=b+4 \end{gathered}$$
Hence, $b-5$ and $b+4$ are the roots of the given equation.