Question

Solve the following quadratic equation by completing square method
$x^2+10x+21=0$

Answer 1

we have, $x^2+10x+21=0$

$\therefore x^2+10x=-21$ ....... $\left(i\right)$

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

$={(\frac{1}{2}\times 10)}^2$

$=25$

Adding $25$ to both sides of equation $\left(i\right)$, we get

$\therefore x^2+10x+25=-21+25$

$\therefore (x+5{)}^2=4$

$\therefore (x+5{)}^2=(2{)}^2$

Taking square roots on both sides
$x+5=\pm 2$

$x=-5\pm 2$

$x=-5+2=-3$ or $x=-5-2=-7$

Hence $-3$ and $-7$ are the roots of the given quadratic equation