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Question: If m₁ and m₂ are roots of the equation x² + (√3 + 2) x + √3 - 1 = 0, then the area of A formed by th...

If m₁ and m₂ are roots of the equation x² + (√3 + 2) x + √3 - 1 = 0, then the area of A formed by the line y = m₁ x, y = m₂x, y = c is

A

33+114c2\frac{\sqrt{33} + \sqrt{11}}{4} c^2

B

32+1116c\frac{\sqrt{32} + \sqrt{11}}{16} c

C

33+104c2\frac{\sqrt{33} + \sqrt{10}}{4} c^2

D

33+214c3\frac{\sqrt{33} + \sqrt{21}}{4} c^3

Answer

33+114c2\frac{\sqrt{33} + \sqrt{11}}{4} c^2

Explanation

Solution

The area of the triangle formed by the lines y=m1xy = m_1x, y=m2xy = m_2x, and y=cy = c is given by c22∣m1−m2m1m2∣\frac{c^2}{2} \left| \frac{m_1 - m_2}{m_1 m_2} \right|. From Vieta's formulas for x2+(3+2)x+3−1=0x^2 + (\sqrt{3} + 2) x + \sqrt{3} - 1 = 0: m1+m2=−(3+2)m_1 + m_2 = -(\sqrt{3} + 2) m1m2=3−1m_1 m_2 = \sqrt{3} - 1 We calculate (m1−m2)2=(m1+m2)2−4m1m2(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2: (m1−m2)2=(−(3+2))2−4(3−1)=(7+43)−(43−4)=11(m_1 - m_2)^2 = (-(\sqrt{3} + 2))^2 - 4(\sqrt{3} - 1) = (7 + 4\sqrt{3}) - (4\sqrt{3} - 4) = 11. So, ∣m1−m2∣=11|m_1 - m_2| = \sqrt{11}. Substituting these values into the area formula: Area =c22∣113−1∣=c2112(3−1)= \frac{c^2}{2} \left| \frac{\sqrt{11}}{\sqrt{3} - 1} \right| = \frac{c^2 \sqrt{11}}{2(\sqrt{3} - 1)}. Rationalizing the denominator: Area =c2112(3−1)×3+13+1=c211(3+1)2(3−1)=c2(33+11)4= \frac{c^2 \sqrt{11}}{2(\sqrt{3} - 1)} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{c^2 \sqrt{11}(\sqrt{3} + 1)}{2(3 - 1)} = \frac{c^2 (\sqrt{33} + \sqrt{11})}{4}.