The problem at hand is, find the solutions of x in the following equation:

(x2−7x+11)x2−7x+6=1

My friend who gave me this questions, told me that you can find 6 solutions without needing to graph the equation.

My approach was this: Use factoring and the fact that z0=1 for z≠0 and 1z=1 for any z.

Factorising the exponent, we have:

x2−7x+6=(x−1)(x−6)

Therefore, by making the exponent = 0, we have possible solutions as x∈{1,6}

Making the base of the exponent = 1, we get x2−7x+10=0

(x−2)(x−5)Hence we can say x∈{2,5}.

However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?

**Answer**

Denote a=x2−7x+11. The equation becomes aa−5=1, or equivalently* aa=a5, which has in R the solutions a∈{5,1,−1}. Solving the corresponding quadratic equations we get the solutions x∈{1,6,2,5,3,4}.

***Note added:** a=0 is excluded in both equations.

**Attribution***Source : Link , Question Author : Community , Answer Author : user376343*