If the sum of the squares of the roots of x^2-(-2)x-(1+)=0 is least, then is equal to

Question

If the sum of the squares of the roots of $x^2-(\sin\alpha-2)x-(1+\sin\alpha)=0$ is least, then $\alpha$ is equal to

Accepted Answer

$x_1+x_2=\sin\alpha-2$, $x_1x_2=-(1+\sin\alpha)$.

$$x_1^2+x_2^2=\sin^2\alpha-2\sin\alpha+6,$$

which (checking the candidate options) is minimized at $\alpha=\dfrac\pi2$.

Solutions · 1