If m_1 and m_2 are the roots of the equation | vedclass

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(B) Given the quadratic equation $x^2+(\sqrt{3}+2) x+(\sqrt{3}-1)=0$. By Vieta's formulas,the sum of roots $m_1+m_2 = -(\sqrt{3}+2)$ and the product o

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$\left(\frac{\sqrt{33}+\sqrt{11}}{4}\right) \cdot c^2$

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(B) Given the quadratic equation $x^2+(\sqrt{3}+2) x+(\sqrt{3}-1)=0$. By Vieta's formulas,the sum of roots $m_1+m_2 = -(\sqrt{3}+2)$ and the product of roots $m_1 m_2 = \sqrt{3}-1$. The difference of the roots is $|m_1-m_2| = \sqrt{(m_1+m_2)^2 - 4m_1 m_2} = \sqrt{(\sqrt{3}+2)^2 - 4(\sqrt{3}-1)} = \sqrt{3+4+4\sqrt{3}-4\sqrt{3}+4} = \sqrt{11}$. The vertices of the triangle formed by $y=m_1 x, y=m_2 x$ and $y=c$ are $(0,0), (c/m_1, c)$ and $(c/m_2, c)$. The area of the triangle is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = \frac{1}{2} |0(c-c) + \frac{c}{m_1}(c-0) + \frac{c}{m_2}(0-c)| = \frac{1}{2} c^2 |\frac{1}{m_1} - \frac{1}{m_2}| = \frac{1}{2} c^2 \frac{|m_2-m_1|}{|m_1 m_2|}$. Substituting the values,Area $= \frac{1}{2} c^2 \frac{\sqrt{11}}{\sqrt{3}-1} = \frac{1}{2} c^2 \frac{\sqrt{11}(\sqrt{3}+1)}{3-1} = \frac{\sqrt{33}+\sqrt{11}}{4} c^2$.

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 $y=c$ is:

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