If $m_1$ and $m_2$ are the roots of the equation $x^2+\left(\sqrt{3}+2\right)x+\left(\sqrt{3}-1\right)=0$, then the area of the triangle formed by the lines $y=m_1x,y=m_2x$ and $y=2$ is :

The lines y equals m1 times x and y equals m2 times x intersect the line y equals 2 at the points where x equals 2 divided by m1 and x equals 2 divided by m2. The base of the triangle on the line y equals 2 has length 2 divided by m1 minus 2 divided by m2, and the height is 2. Using Vieta's formulas, 1 divided by m1 plus 1 divided by m2 equals the negative coefficient of x, giving negative 2 minus the square root of 3, and 1 divided by m1 times 1 divided by m2 equals 1 divided by the square root of 3 minus 1. The area is half the base times the height, which simplifies to the square root of 33 minus the square root of 11. The result is sqrt 33 minus sqrt 11 sq. units.

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