0
$\begingroup$

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.)

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

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

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 $0$: $$x^2-7x+11=0$$ $$x_{1/2}=\frac{7\pm\sqrt{7^2-4\cdot1\cdot11}}{2}=\frac{7\pm\sqrt{5}}{2}$$
$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

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.