What is the sum of all real numbers $x$ such that $(x^2-5x+5)^{(x^2-7x+12)}=1$?

So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up.

Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.

Solving $x^2-5x+5=1$ for $1^x=1$ gives $x=1, 4$.

Adding them up gives $1+3+4=8$.

This is wrong. What did I do wrong? Did I miss a case? If so, what case have I missed?



  • $\begingroup$ No you don't. When you put $x=4$ you get $1^0$. The $0^0$ ambiguity is not the OP's problem here, see the answers. $\endgroup$ – Oscar Lanzi Mar 22 at 23:05


You have missed $$(-1)^{\text{even number}}=1$$

Now if $x^2-5x+5=-1,x=?$

Which values of $x$ make the exponent even?

  • $\begingroup$ Oh! I see! Thank you very much! $\endgroup$ – Max0815 Mar 22 at 2:12
  • $\begingroup$ @lab check my edit. I wanted to clarify that we can have any even exponent on $-1$. The words are rendered by using \text{words} between the dollar signs. $\endgroup$ – Oscar Lanzi Mar 22 at 2:15
  • 1
    $\begingroup$ @Oscar, Thanks for the update $\endgroup$ – lab bhattacharjee Mar 22 at 2:24

Three cases: $$x^0=1$$ $$1^x=1$$ $$(-1)^{\text{even #}}=1$$ Substituting into the original equation gives $$x=3, 4$$ $$x=1, 4$$ $$x=2, 3$$ We don't add in the solutions already accounted for, namely, 3 and 4. $\sum=3+4+1+2=10$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.