Question

If $m_1$ and $m_2$ are the roots of the equation $x^2+(\sqrt{3}+2)x+(\sqrt{3}-1)=0$, then the area of the triangle formed by the lines $y=m_{1}x,y=m_{2}x$ and $y=2$ is :

• A. $\sqrt{33}-\sqrt{11}$ sq. units
• B. $\sqrt{11}+\sqrt{33}$ sq. units
• C. $2\sqrt{33}$ sq. units
• D. $121$ sq. units

Answer 1

The correct option is A $\sqrt{33}-\sqrt{11}$ sq. units

Given:$m_1$ and $m_2$ are the roots of the equation

$x^2+(\sqrt{3}+2)x+(\sqrt{3}-1)=0$

$\therefore$ value of $m_1$ and $m_2$ is found out by quadratic formula as

$m_1=\frac{-(\sqrt{3}+2)+\sqrt{{(\sqrt{3}+2)}^2-4(\sqrt{3}-1)}}{2}$

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

$=\frac{-(\sqrt{3}+2)+\sqrt{11}}{2}$

$\Rightarrow \left|m_1\right|=|-\left[\frac{(\sqrt{3}+2)-\sqrt{11}}{2}\right]|=\frac{(\sqrt{3}+2)-\sqrt{11}}{2}$

and $\left|m_2\right|=|-\left[\frac{(\sqrt{3}+2)+\sqrt{11}}{2}\right]|=\frac{(\sqrt{3}+2)+\sqrt{11}}{2}$

Area of triangle formed $=$ Area of shaded region

$=\frac{1}{2}\times$ base $\times$ height

Where 'c' is defined as the perpendicular height

$=\frac{1}{2}\times [\frac{c}{\left|m_1\right|}-\frac{c}{\left|m_2\right|}]\times c$

$=\frac{1}{2}{c}^2[\frac{1}{\left|m_1\right|}-\frac{1}{\left|m_2\right|}]$

$=\frac{1}{2}{c}^2[\frac{2}{\sqrt{3}+2-\sqrt{11}}-\frac{2}{\sqrt{3}+2+\sqrt{11}}]$

$=c^2[\frac{1}{\sqrt{3}+2-\sqrt{11}}-\frac{1}{\sqrt{3}+2+\sqrt{11}}]$

$=c^2\left[\frac{\sqrt{3}+2+\sqrt{11}-\sqrt{3}-2+\sqrt{11}}{{(\sqrt{3}+1)}^2-{\left(\sqrt{11}\right)}^2}\right]$

$=c^2\left[\frac{2\sqrt{11}}{4\sqrt{3}-4}\right]=c^2\left[\frac{\sqrt{11}}{\sqrt{3}-1}\right]$

$=\frac{11c^2}{2\sqrt{11}\times (\sqrt{3}-1)}=\frac{11c^2}{2(\sqrt{33}-\sqrt{11})}\times \frac{\sqrt{33}+\sqrt{11}}{\sqrt{33}+\sqrt{11}}$

$=\frac{11c^2(\sqrt{33}+\sqrt{11})}{2(33-11)}=\frac{11c^2(\sqrt{33}+\sqrt{11})}{2\times 22}$

$=\frac{c^2(\sqrt{33}+\sqrt{11})}{4}$