Question

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)`?

Answer 1

`(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.