Question
Solve the following equations by using the method of completing the square:
$ 7 x^2+3 x-4=0 $
$ 7 x^2+3 x-4=0 $
✓
Answer
$ 7 x^2+3 x-4=0 $
$ \Rightarrow 49 x^2+21 x-28=0(\text { Multiplying both sides by } 7)$
$ \Rightarrow 49 x^2+21 x=28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or $x = -1$
Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.
$ \Rightarrow 49 x^2+21 x-28=0(\text { Multiplying both sides by } 7)$
$ \Rightarrow 49 x^2+21 x=28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or $x = -1$
Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.
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