Question

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

Answer 1

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

$\Rightarrow$ $x^2+\frac{3}{7}x-\frac{4}{7}=0$

Here. $a=1,b=\frac{3}{7},c=-\frac{4}{7}$

$\Rightarrow$ $x^2+\frac{3}{7}x=\frac{4}{7}$

Now, ${\left(\frac{b}{2}\right)}^2={\left(\frac{3}{7\times 2}\right)}^2=\frac{9}{196}$

Adding $\frac{9}{196}$, on both sides of equation,

$\Rightarrow$ $x^2+\frac{3}{7}x+\frac{9}{196}=\frac{4}{7}+\frac{9}{196}$

$\Rightarrow$ ${(x+\frac{3}{14})}^2=\frac{112+9}{196}$

$\Rightarrow$ ${(x+\frac{3}{14})}^2=\frac{121}{196}$

Taking square roots on both sides,

$\Rightarrow$ $x+\frac{3}{14}=\pm \frac{11}{14}$

$\Rightarrow$ $x=\frac{11}{14}-\frac{3}{14}$ and $x=-\frac{11}{14}-\frac{3}{14}$

$\therefore$ $x=\frac{4}{7}$ and $x=-1$