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

Answer:

#(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.