Question

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

$7x^2+3x-4=0$

Answer 1

$$\begin{gathered} 7x^2+3x-4=0 \\ \Rightarrow 49x^2+21x-28=0\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}7\right) \\ \Rightarrow 49x^2+21x=28 \\ \Rightarrow {\left(7x\right)}^2+2\times 7x\times \frac{3}{2}+{\left(\frac{3}{2}\right)}^2=28+{\left(\frac{3}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{3}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(7x+\frac{3}{2}\right)}^2=28+\frac{9}{4}=\frac{121}{4}={\left(\frac{11}{2}\right)}^2 \\ \Rightarrow 7x+\frac{3}{2}=\pm \frac{11}{2}\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow 7x+\frac{3}{2}=\frac{11}{2}\mathrm{or}7x+\frac{3}{2}=-\frac{11}{2} \\ \Rightarrow 7x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4\mathrm{or}7x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7 \end{gathered}$$
$\Rightarrow x=\frac{4}{7}\mathrm{or}x=-1$
Hence, $\frac{4}{7}$ and −1 are the roots of the given equation.