Question

If two sides of a triangle are the roots of the equation $4x^2-\left(2\sqrt{6}\right)x+1=0$ and the included angle is ${60}^o$, then the third side is

• A. $\sqrt{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $2\sqrt{3}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

The roots of the given equation are given by
$x=\frac{2\sqrt{6}\pm \sqrt{24-16}}{8}$

$=\frac{2\sqrt{6}\pm 2\sqrt{2}}{8}=\frac{\sqrt{6}\pm \sqrt{2}}{4}$

Now if a represents the third side, then
$\cos {60}^o=\frac{{\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}^2+\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)-a^2}{2\times \frac{\sqrt{6}+\sqrt{2}}{4}\times \frac{\sqrt{6}-\sqrt{2}}{4}}$

$\Rightarrow \frac{1}{2}=\frac{2(6+2)-16a^2}{2(6-2)}$

$\Rightarrow a^2=\frac{3}{4}\Rightarrow a=\frac{\sqrt{3}}{2}$.