Question

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

$2x^2+5x-3=0$

Answer 1

$$\begin{gathered} 2x^2+5x-3=0 \\ \Rightarrow 4x^2+10x-6=0\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2\right) \\ \Rightarrow 4x^2+10x=6 \\ \Rightarrow {\left(2x\right)}^2+2\times 2x\times \frac{5}{2}+{\left(\frac{5}{2}\right)}^2=6+{\left(\frac{5}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{5}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(2x+\frac{5}{2}\right)}^2=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}={\left(\frac{7}{2}\right)}^2 \\ \Rightarrow 2x+\frac{5}{2}=\pm \frac{7}{2}\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow 2x+\frac{5}{2}=\frac{7}{2}\mathrm{or}2x+\frac{5}{2}=-\frac{7}{2} \\ \Rightarrow 2x=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1\mathrm{or}2x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6 \end{gathered}$$
$\Rightarrow x=\frac{1}{2}\mathrm{or}x=-3$
Hence, $\frac{1}{2}$ and $-3$ are the roots of the given equation.