# Quadratic equation with indicies as another quadratic.

I'm trying to solve the following question below (Please do excuse the formatting)... $$x^{x^2-7x+11} = 1$$ Now, so far, I have calculated that as $1 =x^0$ that I can form an equation which is

$$x^2 - 7x+11 = 0$$ and the values of x that it gives are $x = 5$ and $x = 6$. However, when graphing this solution, I also get the result of $x = -1$ and $x = 1$.

How is this possible (in an algebraic matter)?

Thanks.

The results were checked with grapher and wolfram alpha.

(P.S. Any formatting to the quadratic notation of $x^2$ and the quadratic expressions would be grately appreciated.)

• When $x=1$ you have $1^{1-7+11}=1^5=1$. – vadim123 Sep 12 '15 at 19:52
• your roots are wrong. Plug them into the quadratic to verify. And x=1 is trivial. – MrYouMath Sep 12 '15 at 19:53
• Note that plugging in shows that $x=-1$ does not work. – André Nicolas Sep 12 '15 at 19:56
• I do apologise @vadim123 but I do not understand. Can a full answer maybe solve the problem please? – vik1245 Sep 12 '15 at 19:58

## 2 Answers

$$x^{x^2-7x+11} = 1$$

This expression can be equal to one,

1. Base: If $x = 1$, as $1^{x^2-7x+11}=1$
2. Base: Check if $x = -1$ has a positive power? That is not the case, as $x = -1$ has power $19$. Hence, $x=-1$ can't be a solution.
3. Power is equal to $0$: $$x^2-7x+11=0$$ $$x_{1/2}=\frac{7\pm\sqrt{7^2-4\cdot1\cdot11}}{2}=\frac{7\pm\sqrt{5}}{2}$$

so far, I have calculated that as $1=x^0$

That's not the only way to get to the result of 1.

Recall that $$\overbrace{1\cdot 1\cdot 1\cdot 1 ... \cdot 1\cdot 1}^{\text{n times}}=1$$ and $$\overbrace{-1\cdot -1\cdot -1\cdot -1 ... \cdot -1\cdot -1}^{\text{2n times}}=1$$

That means that you should also find solutions for $x =-1$ and $x=1$ which create solutions in the form of

$$1=(-1)^{2n}$$ and $$1=1^{n}$$

with $x =-1$, that is

$$(-1)^2 - 7(-1)+11 = 2n$$

and $x =1$, that is

$$1^2 - 7\cdot 1+11 = n$$