If $\cos^4\theta+\alpha$ and $\sin^4\theta+\alpha$ are the roots of the equation $x^2+4x+2=0$, then values of $\alpha$ are
- $2$
- $-1$
- $-2$
-
None of these
The roots are cos^4(theta)+alpha and sin^4(theta)+alpha. Their sum is 1+2alpha = -4 (so alpha = -2.5) and their product is (cos^4(theta)+alpha)(sin^4(theta)+alpha) = 2. Substituting alpha = -2.5 leads to a contradiction, as the values of sin and cos are bounded. Thus, no real alpha satisfies the equation.
From the given quadratic equation, the sum and product of the roots are (cos^4 theta + alpha) + (sin^4 theta + alpha) = -4 and (cos^4 theta + alpha)(sin^4 theta + alpha) = 2. Using the identity a^2 + b^2 = (a + b)^2 - 2ab, we substitute the known sum and product to get cos^4 theta + sin^4 theta + 2 alpha = -2. Applying the reduction identity cos^4 theta + sin^4 theta = 1 - 2 sin^2 theta cos^2 theta results in the equation 2 alpha^2 + 5 alpha + 1 = 0. Solving this quadratic equation for alpha yields two irrational roots, neither of which equals 2, -1, or -2, so the correct choice is None of these.