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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

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  • $\begingroup$ 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, 2019 at 23:05

2 Answers 2

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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?

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  • $\begingroup$ Oh! I see! Thank you very much! $\endgroup$
    – Max0815
    Mar 22, 2019 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, 2019 at 2:15
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    $\begingroup$ @Oscar, Thanks for the update $\endgroup$
    – lab bhattacharjee
    Mar 22, 2019 at 2:24
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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. $\sum=3+4+1+2=10$.

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