We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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.

 Aug 20, 2019

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


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


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


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


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

 Aug 20, 2019

21 Online Users