For #a,b,c# non- zero, real distinct, the equation, #( a^2+b^2)x^2 -2b(a+c) x +b^2+c^2=0# has non-zero real roots. One of these roots is also the root of the equation? A) #a^2x^2-a(b-c)x+bc=0# B) #a^2x^2+a(c-b)x-bc=0# C) #(b^2+c^2)x^2-2a(b+c)x+a^2=0#

1 Answer
Mar 5, 2017

Answer:

See below.

Explanation:

Solving for #x# in #( a^2+b^2)x^2 -2b(a+c) x +b^2+c^2=0# we have

#x=(b (a + c) - sqrt[-(b^2 - a c)^2])/(a^2 + b^2)# The solutions must be real so necessarily #b^2 - a c=0# and the roots are real identical

#x = {c/sqrt(ac),c/sqrt(ac)}#

After substituting #b = pmsqrt(ac)# into the equations from A) to C)

we obtain for the option B) the roots

#x = {c/sqrt(ac),c/a}#

so the correct option is B)