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 - b²)(x²+1)#?

1 Answer
Jul 27, 2016

Answer:

The roots are complex conjugate

Explanation:

If the roots of

#a x^2 + 2 b x + c=0#

are real and distint then #b^2-ac>0#

Now grouping

#(a + c) (a x^2 + 2 b x + c) - 2 (a c - b^2) (x^2 + 1)=0#

we have

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

and solving for #x#

#x = (-b (a + c) + sqrt[(4 b^2 + (a - c)^2) (a c-b^2)])/( a^2 + 2 b^2 - a c)#

and #a c-b^2<0# so the roots are complex conjugate