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Question

# Solve each of the following equations by using the method of completing the square: $7{x}^{2}+3x-4=0$

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Solution

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

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