If -i and alpha are the roots of the equation | vedclass

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(A) Given the quadratic equation $iz^2 - 2(i+1)z + (2-i) = 0$. Since $-i$ is a root,the sum of roots is $\alpha + (-i) = -\frac{b}{a} = \frac{2(i+1)}{

Vedclass · Accepted Answer

$-117$

Vedclass · Answer

(A) Given the quadratic equation $iz^2 - 2(i+1)z + (2-i) = 0$. Since $-i$ is a root,the sum of roots is $\alpha + (-i) = -\frac{b}{a} = \frac{2(i+1)}{i} = 2(1-i) = 2-2i$. Thus,$\alpha = 2-i$. However,the question asks for the value of $5^3 \cos 6	heta$ based on $	an 	heta = -\frac{1}{2}$. Using the formula $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta} = \frac{3(-1/2) - (-1/8)}{1 - 3(1/4)} = \frac{-3/2 + 1/8}{1/4} = \frac{-11/8}{1/4} = -\frac{11}{2}$. Now,$5^3 \cos 6	heta = 125 \left( \frac{1 - 	an^2 3	heta}{1 + 	an^2 3	heta} ight) = 125 \left( \frac{1 - (-11/2)^2}{1 + (-11/2)^2} ight) = 125 \left( \frac{1 - 121/4}{1 + 121/4} ight) = 125 \left( \frac{-117/4}{125/4} ight) = -117$.

If $-i$ and $\alpha$ are the roots of the equation $iz^2 - 2(i+1)z + (2-i) = 0$,$\tan \theta = \frac{-1}{2}$ and $\theta \in 4^{\text{th}}$ quadrant,then $5^3 \cos 6\theta =$

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