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

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(A) For the first equation $x^2 + 2bx + b = 0$,the roots are $r_1 = \cos^4 	heta + \alpha$ and $r_2 = \sin^4 	heta + \alpha$.The difference of the r

Vedclass · Accepted Answer

$2$

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(A) For the first equation $x^2 + 2bx + b = 0$,the roots are $r_1 = \cos^4 	heta + \alpha$ and $r_2 = \sin^4 	heta + \alpha$.The difference of the roots is $|r_1 - r_2| = |(\cos^4 	heta + \alpha) - (\sin^4 	heta + \alpha)| = |\cos^4 	heta - \sin^4 	heta| = |(\cos^2 	heta - \sin^2 	heta)(\cos^2 	heta + \sin^2 	heta)| = |\cos 2	heta|$.From the quadratic formula,the difference of roots is $\sqrt{D}/a = \sqrt{(2b)^2 - 4(1)(b)} = \sqrt{4b^2 - 4b} = 2\sqrt{b^2 - b}$.Thus,$|\cos 2	heta| = 2\sqrt{b^2 - b}$.For the second equation $x^2 + 4x + 2 = 0$,the roots are $s_1 = \cos^2 	heta + \beta$ and $s_2 = \sin^2 	heta + \beta$.The difference of the roots is $|s_1 - s_2| = |(\cos^2 	heta + \beta) - (\sin^2 	heta + \beta)| = |\cos^2 	heta - \sin^2 	heta| = |\cos 2	heta|$.From the quadratic formula,the difference of roots is $\sqrt{D}/a = \sqrt{4^2 - 4(1)(2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.Equating the two expressions for $|\cos 2	heta|$:$2\sqrt{b^2 - b} = 2\sqrt{2}$$\sqrt{b^2 - b} = \sqrt{2}$$b^2 - b = 2$$b^2 - b - 2 = 0$$(b - 2)(b + 1) = 0$So,$b = 2$ or $b = -1$. Since $b=2$ is an option,the correct answer is $2$.

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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