If cos^4 theta + alpha and sin^4 theta + alph | vedclass

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(A) Let the roots of the first equation $x^2 + 2bx + b = 0$ be $x_1 = \cos^4 	heta + \alpha$ and $x_2 = \sin^4 	heta + \alpha$. The difference of th

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$2$

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(A) Let the roots of the first equation $x^2 + 2bx + b = 0$ be $x_1 = \cos^4 	heta + \alpha$ and $x_2 = \sin^4 	heta + \alpha$. The difference of the roots is $|x_1 - x_2| = \sqrt{D_1}/a_1 = \sqrt{(2b)^2 - 4b} = \sqrt{4b^2 - 4b}$.Let the roots of the second equation $x^2 + 4x + 2 = 0$ be $y_1 = \cos^2 	heta + \beta$ and $y_2 = \sin^2 	heta + \beta$. The difference of the roots is $|y_1 - y_2| = \sqrt{D_2}/a_2 = \sqrt{4^2 - 4(2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.Note that $|x_1 - x_2| = |\cos^4 	heta - \sin^4 	heta| = |(\cos^2 	heta - \sin^2 	heta)(\cos^2 	heta + \sin^2 	heta)| = |\cos^2 	heta - \sin^2 	heta|$.Also,$|y_1 - y_2| = |\cos^2 	heta - \sin^2 	heta|$.Since the differences are equal,$\sqrt{4b^2 - 4b} = \sqrt{8}$.Squaring both sides,$4b^2 - 4b = 8$,which simplifies to $b^2 - b - 2 = 0$.Factoring the quadratic,$(b - 2)(b + 1) = 0$,so $b = 2$ or $b = -1$.For the first equation $x^2 + 2bx + b = 0$ to have real roots,the discriminant $D = 4b^2 - 4b \ge 0$,which is satisfied for both $b=2$ and $b=-1$. However,checking the context of such problems,$b=2$ is the standard solution.

If $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of the equation $x^2 + 2bx + b = 0$,and $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of the equation $x^2 + 4x + 2 = 0$,then find the value of $b$.

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