Sender: David Xiao
Subject: Math grade 12 ( quadratic equation )

Hi I have a problem here which I do not know how to solve it . Please help me thanks .

Find the value of a such that 4x² + 4(a-2)x - 8a² + 14a + 31 = 0 has real roots whose sum of squares is minimum.

Thank you

Consider the general quadratic Ax² + Bx + C. If the roots are real then the discriminant, B² - 4AC must be non-negative. In your example A = 4, B = 4(a - 2) and C = - 8a² + 14a + 31 and the discriminant is 144(a - 3)(a + 1). Thus the discriminant is non-negative if and only if a >= 3 or a <= -1.

Suppose that the roots are p and q, then p + q = -B/A and pq = C/A. Thus

10a² - 22a - 23 is a quadratic with minimum at 22/(20) = 1.1. Thus this quadratic is decreasing for a less than 1.1 and increasing for a greater than 1.1. Hence the minimum is at either a = -1 or a = 3.