If the roots of the equation ax^2+2bx+c=0 are real and distinct then find the nature of the roots of the equation (a+c)(ax^2+2bx+c) = 2(ac - b^2)(x^2+1)?

1 Answer
Jun 19, 2016

(a+c)(ax^2+2bx+c) = 2(ac-b^2)(x^2+1)

has a Complex conjugate pair of non-Real zeros.

Explanation:

Discriminant

The discriminant Delta of a quadratic expression Ax^2+Bx+C is B^2-4AC.

The sign of the discriminant determines the kind of zeros the quadratic has:

  • Delta > 0 : Two distinct Real zeros.
  • Delta = 0 : One repeated Real zero.
  • Delta < 0 : A Complex conjugate pair of non-Real zeros.

Solution

The discriminant of ax^2+2bx+c is:

Delta_1 = (2b)^2-4ac = 4(b^2-ac)

Since ax^2+2bx+c=0 has distinct Real zeros, Delta_1 > 0, hence:

b^2-ac > 0

Let k = b^2-ac > 0

Given:

(a+c)(ax^2+2bx+c) = 2(ac-b^2)(x^2+1)

Expand and rearrange into standard form:

(a^2-ac+2b^2)x^2+2b(a+c)x+(c^2-ac+2b^2) = 0

This has discriminant:

Delta_2 = (2b(a+c))^2-4(a^2-ac+2b^2)(c^2-ac+2b^2)

=4(b^2(a+c)^2-(a^2-ac+2b^2)(c^2-ac+2b^2))

=4((ac+k)(a+c)^2-(a^2+ac+2k)(c^2+ac+2k))

=4(color(red)(cancel(color(black)(ac(a+c)^2)))+k(a+c)^2-color(red)(cancel(color(black)(ac(a+c)^2)))-2k(a+c)^2-4k^2)

=4(-k(a+c)^2-4k^2) < 0

So:

(a+c)(ax^2+2bx+c) = 2(ac-b^2)(x^2+1)

has a Complex conjugate pair of non-Real zeros.