This is a problem originally published on Problem of the Week of University of Waterloo. The question was about finding the roots of a functions. I thought it would be really easy since Desmos seems to be able to solve all functions, however in this interesting question it fails, which is why I am posting it here.

Original Question

You may be surprised to learn that the equation $(x^2-5x+5)^{x^2+4x-60}=1$ has five solutions.

Determine all five values of $x$ that satisfy the equation.

Attempts

At first I was trying to solve this as a polynomial, moving “1” from the RHS to the LHS, but then I did not know what to do. This is way more complicated to solve and I did not even know where to start.

Failing to solve this on my own, I put the whole function into Desmos, and this is what it gives me:

Ok here I can see four roots, but where is the fifth? Is it a real number even? And also what is this?

Eventually I asked my tutor about this and when he pointed me the way out I realised that I was being really dumb.

So in this functions there are three cases that will results in the RHS being 1.

Case #1

index ​$=x^2+4x-60=0$
$(x+10)(x-6)=0$
$​x=6$ or $x=-10$

Case #2

base = $x^2-5x+5=1$
$x^2-5x+4=0$
$(x−4)(x−1)=0$
$x=1$ or $x=4$

Case #3

This is the root that is not showing in Desmos.

base = -1, index = even
So $x^2+4x-60$ must be a even number
base = -1, $x^2-5x+5=-1$
$x^2-5x+6=0$
$(x−2)(x−3)=0$
$x=2$ or $x=3$
when $x=2$, index = $2^2+4*2-60=4+8-60=-48$
when $x=3$, index = $3^2+4*3-60=9+12-60=-39$ ≠ even
$x=2$

Answer

So the final 5 roots?
$x=-10, 1, 2, 4, 6$

Hope you find this interesting!