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
has a Complex conjugate pair of non-Real zeros.
Explanation:
Discriminant
The discriminant
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
Delta_1 = (2b)^2-4ac = 4(b^2-ac)
Since
b^2-ac > 0
Let
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.