+0

0
54
1

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

#1
+5766
+1

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

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

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

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

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

.
Aug 20, 2019