# Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$.

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?

Thanks!

Max0815

• 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. – Oscar Lanzi Mar 22 at 23:05

Hint

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

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

Which values of $$x$$ make the exponent even?

• Oh! I see! Thank you very much! – Max0815 Mar 22 at 2:12
• @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. – Oscar Lanzi Mar 22 at 2:15
• @Oscar, Thanks for the update – 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. $$\sum=3+4+1+2=10$$.