Question

Solve the following quadratic equation by completing the square method:
$m^2-3m-1=0$

Answer 1

$m^2-3m-1=0$
$m^2-3m=1$ $\to$ (1)
Third term $={\left(\frac{1}{2}coeeficientofm\right)}^2$
${(\frac{1}{2}\times (-3\left)\right)}^2={\left(\frac{-3}{2}\right)}^2=\frac{9}{4}$
Adding $\frac{9}{4}$ to both sides of equation (1), we get
$m^2-3m+\frac{9}{4}=1+\frac{9}{4}$
$\therefore$ $m^2-3m+\frac{9}{4}=\frac{4+9}{4}$
$\therefore$ ${(m-\frac{3}{2})}^2=\frac{13}{4}$
Taking square roots on both sides
$\therefore$ $m-\frac{3}{2}=\pm \frac{\sqrt{13}}{2}$
$\therefore$ $m=\frac{3}{2}+\frac{\sqrt{13}}{2}$ or $m=\frac{3}{2}-\frac{\sqrt{13}}{2}$
$m=\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}$ are the roots of the given quadratic equation.