Question

Solve the following quadratic equation by completing square method
$x^2+11x+30=0$

Answer 1

$x^2+11x+30=0$

$\therefore x^2+11x=-30$ .......... $\left(i\right)$

Third term $={\left(\frac{1}{2}\text{coefficient of}x\right)}^2$

$={(\frac{1}{2}\times 11)}^2$

$=\frac{121}{4}$

Adding $\frac{121}{4}$ to both sides of equation (i), we get

$x^2+11x+\frac{121}{4}=-30+\frac{121}{4}$

$\therefore {(x+\frac{11}{2})}^2=\frac{1}{4}$

$\therefore {(x+\frac{11}{2})}^2={\left(\frac{1}{2}\right)}^2$

Taking square roots on both sides.

$x+\frac{11}{2}=\pm \frac{1}{2}$

$x=-\frac{11}{2}\pm \frac{1}{2}$

$x=-\frac{11}{2}+\frac{1}{2}=\frac{-10}{2}=-5$

or $x=\frac{-11}{2}-\frac{1}{2}=\frac{-12}{2}=-6$

Hence $-5$ and $-6$ are the roots of the given quadratic equation