Find the two real roots of the equation x^4 - 5 = 4x^2 . Give the roots in an exact form. [4]
x4 - 5 = 4x2. In order to find the roots of this equation we must factorise it, and, to do so, it must be rearranged such that all the terms are on one side. Therefore, subtract 4x2 from both sides to obtain x4 - 4x2 - 5 = 0.This equation can now be seen to be a quadratic equation in the form of x2, and so must be factorised like any other equation. The integers that multiply together to give -5 and sum together to make -4 are 5 and -1. Therefore this quadratic can be factorised to give (x2-5)(x2+1) = 0. As x2+1=0 has no REAL solutions, the only bracket worth considering is (x2-5) = 0. This gives the answers of x = +/- root 5.Alternative method: substitute u = x2 into the equation, to give u2 - 4u - 5 =0. This might make it easier to spot that the equation is a quadratic. Factorise in the same way, to give (u-5)(u+1) = 0. u = -1 and u =5. However, as u = x2, and x2 = -1 has no real solutions, again this can be discarded.
