Question

Solve each of the following quadratic equations:

$9x^2-3x-2=0$

Answer 1

We write, $-3x=3x-6x$ as $9x^2\times \left(-2\right)=-18x^2=3x\times \left(-6x\right)$

$$\begin{gathered} \therefore 9x^2-3x-2=0 \\ \Rightarrow 9x^2+3x-6x-2=0 \\ \Rightarrow 3x\left(3x+1\right)-2\left(3x+1\right)=0 \\ \Rightarrow \left(3x+1\right)\left(3x-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x+1=0\mathrm{or}3x-2=0 \\ \Rightarrow x=-\frac{1}{3}\mathrm{or}x=\frac{2}{3} \end{gathered}$$
Hence, the roots of the given equation are $-\frac{1}{3}$ and $\frac{2}{3}$.