Solution to the equation of a polynomial raised to the power of a polynomial.

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

(x27x+11)x27x+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 z0 and 1z=1 for any z.

Factorising the exponent, we have:

x27x+6=(x1)(x6)

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

Making the base of the exponent = 1, we get x27x+10=0
(x2)(x5)

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=x27x+11. The equation becomes aa5=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

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