Question

Solve each of the following quadratic equations:

$x^2+5x-\left(a^2+a-6\right)=0$ [CBSE 2015]

Answer 1

We write, $5x=\left(a+3\right)x-\left(a-2\right)x$ as $x^2\times \left[-\left(a^2+a-6\right)\right]=-\left(a^2+a-6\right)x^2=\left(a+3\right)x\times \left[-\left(a-2\right)x\right]$
$$\begin{gathered} \therefore x^2+5x-\left(a^2+a-6\right)=0 \\ \Rightarrow x^2+\left(a+3\right)x-\left(a-2\right)x-\left(a+3\right)\left(a-2\right)=0 \\ \Rightarrow x\left[x+\left(a+3\right)\right]-\left(a-2\right)\left[x+\left(a+3\right)\right]=0 \\ \Rightarrow \left[x+\left(a+3\right)\right]\left[x-\left(a-2\right)\right]=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+\left(a+3\right)=0\mathrm{or}x-\left(a-2\right)=0 \\ \Rightarrow x=-\left(a+3\right)\mathrm{or}x=a-2 \end{gathered}$$
Hence, $-\left(a+3\right)$ and $\left(a-2\right)$ are the roots of the given equation.