Question

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

Answer 1

$$\begin{gathered} 8x^2-14x-15=0 \\ \Rightarrow 16x^2-28x-30=0\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2\right) \\ \Rightarrow 16x^2-28x=30 \\ \Rightarrow {\left(4x\right)}^2-2\times 4x\times \frac{7}{2}+{\left(\frac{7}{2}\right)}^2=30+{\left(\frac{7}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{7}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(4x-\frac{7}{2}\right)}^2=30+\frac{49}{4}=\frac{169}{4}={\left(\frac{13}{2}\right)}^2 \\ \Rightarrow 4x-\frac{7}{2}=\pm \frac{13}{2}\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow 4x-\frac{7}{2}=\frac{13}{2}\mathrm{or}4x-\frac{7}{2}=-\frac{13}{2} \\ \Rightarrow 4x=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10\mathrm{or}4x=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3 \end{gathered}$$
$\Rightarrow x=\frac{5}{2}\mathrm{or}x=-\frac{3}{4}$
Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.