the roots of the equation

  (3-x)⁴+(2-x)⁴=(5-2x)⁴   are

(A) all real                          (B) all imaginary                     (C) two imaginary and two real

(D) none

Let (3-x) = a & (2-x) = b

Then the equation becomes,

a⁴ + b⁴ = (a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

2ab(2a² + 3ab + 2b²) = 0

Substituting the values of a & b, we get

x = 2,3 or 7x² - 35x +44 = 0

D = 1225 - 1232 = -7 < 0

Hence, (c) 2 imaginary & 2 real roots.

Nice solution Pant.

Let me just add that when you see 2a²+3ab+2b² = 0, you can directly use the discriminant 3²-4.2.2 = 9-16 = -7 and conclude that it has no real solutions. If you have done pairs of straight lines you will recognise this formula. Otherwise divide by b² and then set a/b = x. You will get a quadratic to which one can apply the discriminant formula

(c)  .  as for the solution...well, as pant has already solved it, no use in me solving it again.

um...if i am allowed to ask a question - are there any alternate solutions?