Let $\alpha_ heta$ and $\beta_ heta$ be the distinct roots of $2x^2+(\cos heta)x-1=0$, $ heta\in(0,2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_ heta^4+\beta_ heta^4$, then $16(M+m)$ equals :

Question

Let $\alpha_	heta$ and $\beta_	heta$ be the distinct roots of $2x^2+(\cos	heta)x-1=0$, $	heta\in(0,2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_	heta^4+\beta_	heta^4$, then $16(M+m)$ equals :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\alpha+\beta=-\frac{\cos	heta}2$, $\alpha\beta=-\frac12$, so $\alpha^2+\beta^2=\frac{\cos^2	heta}4+1$ and $\alpha^4+\beta^4=\left(\frac{\cos^2	heta}4+1ight)^2-\frac12$. With $\cos^2	heta\in[0,1]$: $m=1-\frac12=\frac12$ and $M=\frac{25}{16}-\frac12=\frac{17}{16}$. $16(M+m)=17+8=25$. Answer (2).

Q. Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2x^2+(\cos\theta)x-1=0$, $\theta\in(0,2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_\theta^4+\beta_\theta^4$, then $16(M+m)$ equals :

Explanation

Reviewed by Dr. Abubacker

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