Question

Using the method of completing the square, find the roots of $4m^2+4\surd 3m+3=0$.

• A. $(\frac{3}{2}-\sqrt{3}),(-\frac{3}{2}-\sqrt{3})$
• B. $-\frac{3}{2},-\frac{\sqrt{3}}{2}$
• C. $+\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$
• D. $-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D $-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

$4m^2+4\sqrt{3}m+3=0$

Dividing by 4 on both the sides

$m^2+\sqrt{3}m+\frac{3}{4}=0$

$m^2+\sqrt{3}m=-\frac{3}{4}$

Adding the square of one-half of the coefficient of x [i.e., $(\frac{\sqrt{3}}{2}{)}^2$] to both sides

$(m{)}^2+2\times m\times \frac{\sqrt{3}}{2}+(\frac{\sqrt{3}}{2}{)}^2=-\frac{3}{4}+(\frac{\sqrt{3}}{2}{)}^2$

$(m+\frac{\sqrt{3}}{2}{)}^2=-\frac{3}{4}+\frac{3}{4}$

$(m+\frac{\sqrt{3}}{2}{)}^2=0$

$(m+\frac{\sqrt{3}}{2})=0$

$m=-\frac{\sqrt{3}}{2}$

Hence the required roots are $-\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$.