Let tan A, tan B, where A, B ∈ (- π 2, π 2 ), be the roots of the quadratic equation x^2 - 2x - 5 = 0. Then 20sin^2 ( A+B 2 ) is equal to:

Given that $\tan A$tanA and $\tan B$tanB are the roots of the quadratic equation $x^2 - 2x - 5 = 0$x2−2x−5=0.

Sum of the roots: $\tan A + \tan B = 2$tanA+tanB=2

Product of the roots: $\tan A \tan B = -5$tanAtanB=−5

Using the tangent addition formula:

$$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{2}{1 - (-5)} = \dfrac{2}{6} = \dfrac{1}{3}$$tan(A+B)=1−tanAtanBtanA+tanB​=1−(−5)2​=62​=31​

Since $A, B \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$A,B∈(−2π​,2π​) and their product $\tan A \tan B = -5 < 0$tanAtanB=−5<0, one angle is positive and the other is negative. This implies $A+B \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$A+B∈(−2π​,2π​).

Since $\tan(A+B) = \dfrac{1}{3} > 0$tan(A+B)=31​>0, the angle $A+B$A+B must lie in the first quadrant, i.e., $A+B \in \left(0, \dfrac{\pi}{2}\right)$A+B∈(0,2π​). Thus, $\cos(A+B)$cos(A+B) is positive.

$$\cos(A+B) = \dfrac{1}{\sqrt{1 + \tan^2(A+B)}} = \dfrac{1}{\sqrt{1 + \left(\dfrac{1}{3}\right)^2}} = \dfrac{3}{\sqrt{10}}$$cos(A+B)=1+tan2(A+B)​1​=1+(31​)2​1​=10​3​

We need to find the value of $20\sin^2\left(\dfrac{A+B}{2}\right)$20sin2(2A+B​). Using the half-angle identity $\sin^2\theta = \dfrac{1 - \cos(2\theta)}{2}$sin2θ=21−cos(2θ)​:

$$20\sin^2\left(\dfrac{A+B}{2}\right) = 20 \left( \dfrac{1 - \cos(A+B)}{2} \right) = 10(1 - \cos(A+B))$$20sin2(2A+B​)=20(21−cos(A+B)​)=10(1−cos(A+B))

Substituting the value of $\cos(A+B)$cos(A+B):

$$10\left(1 - \dfrac{3}{\sqrt{10}}\right) = 10 - \dfrac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$$10(1−10​3​)=10−10​30​=10−310​

Answer: $10 - 3\sqrt{10}$10−310​