Question

Solve each of the following equations by using the method of completing the square:

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

Answer 1

$$\begin{gathered} 3x^2-x-2=0 \\ \Rightarrow 9x^2-3x-6=0\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}3\right) \\ \Rightarrow 9x^2-3x=6 \\ \Rightarrow {\left(3x\right)}^2-2\times 3x\times \frac{1}{2}+{\left(\frac{1}{2}\right)}^2=6+{\left(\frac{1}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{1}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(3x-\frac{1}{2}\right)}^2=6+\frac{1}{4}=\frac{25}{4}={\left(\frac{5}{2}\right)}^2 \\ \Rightarrow 3x-\frac{1}{2}=\pm \frac{5}{2}\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow 3x-\frac{1}{2}=\frac{5}{2}\mathrm{or}3x-\frac{1}{2}=-\frac{5}{2} \\ \Rightarrow 3x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\mathrm{or}3x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2 \end{gathered}$$
$\Rightarrow x=1\mathrm{or}x=-\frac{2}{3}$
Hence, 1 and $-\frac{2}{3}$ are the roots of the given equation.