Find the value of a such that the quadratic 4x^2 + 4(a-2)x - 8a^2 + 14a + 31 = 0 has real roots, and the sum of their squares is minimized.

Guest Aug 20, 2019

#1**+1 **

Warning: This answer will require you to do significant algebra yourself!

\(\text{real roots means the discriminant is non-negative}\\ \text{the sum of the squares of the roots of $a x^2 + b x+c=0$ is simply}\\ \text{$\dfrac{D+2ac}{a^2}$ where $D=b^2-4ac$ is the discriminant}\)

\(\text{I'm going to rewrite your polynomial to avoid confusion and call it}\\ p(x) = 4x^2 + 4(\alpha-2)x-8\alpha^2+14\alpha + 31\\ a = 4\\ b= 4(\alpha - 2)\\ c=-8\alpha^2+14\alpha +31\)

\(\text{First we need a non-negative discriminant $D$}\\ D=144(-3-2\alpha+\alpha^2)~\text{(I leave you to show this algebra)}\\ D=144(\alpha+1)(\alpha-3)\\ D\geq 0 \Rightarrow \alpha \in (-\infty,-1]\cup [3,\infty)\)

\(\text{Now you want to minimize $S=\dfrac{D+2ac}{a^2}$}\\ \text{subject to $a \in (-\infty, -1]\cup [3, \infty)$}\\~\\ \text{with more algebra you can show}\\ S= 5 \alpha ^2-11 \alpha -\dfrac{23}{2} \\ \text{The vertex of this parabola lies within the excluded interval}\\ \text{so the minimum must be on this boundary}\\ \text{at either $x=-1$ or $x=3$}\\ \left. S\right|_{\alpha = -1} = \dfrac 9 2 \\ \left. S\right|_{\alpha = 3} = \dfrac 1 2 \\ \text{and so the minimum value $\dfrac 1 2$ occurs at $\alpha = 3$}\\ \text{or translating back to your original problem statement that $a=3$}\)

.Rom Aug 20, 2019