Question

Solve each of the following quadratic equations:

$2x^2+x-6=0$

Answer 1

We write, $x=4x-3x$ as $2x^2\times \left(-6\right)=-12x^2=4x\times \left(-3x\right)$

$$\begin{gathered} \therefore 2x^2+x-6=0 \\ \Rightarrow 2x^2+4x-3x-6=0 \\ \Rightarrow 2x\left(x+2\right)-3\left(x+2\right)=0 \\ \Rightarrow \left(x+2\right)\left(2x-3\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow x+2=0\mathrm{or}2x-3=0 \\ \Rightarrow x=-2\mathrm{or}x=\frac{3}{2} \end{gathered}$$
Hence, the roots of the given equation are $-2$ and $\frac{3}{2}$.