Question

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

$3x^2-2x-1=0$

Answer 1

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