# Algebra question resubmited [closed]

Hello I have a quick algebra question.

If I have the following expression

$\large \frac{1}{x^2+1}$, and I multiply the numerator and the denominator by $(x^2+1)$.

Is there any way I can get $x^4+x^2−1$?

The reason I am asking this is because on a problem I did this expression came up and it confused I will post the link.

-

## closed as off-topic by Henry, Nameless, Lost1, Andrew D. Hwang, Sami Ben Romdhane Feb 28 '14 at 0:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Lost1, Andrew D. Hwang, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

This is the question where the expression occurs math.stackexchange.com/questions/260117/… – Fernando Martinez Dec 22 '12 at 15:08
Your previous question was closed. This is a literal copy. Why did you think it wouldn't get closed this time? – akkkk Dec 22 '12 at 15:12
Previous closed question math.stackexchange.com/questions/263482/… – Henry Dec 22 '12 at 15:29
@DavidMitra: only because of the later edits by Crete and amWhy after my comment, and the question is still not clear to me. – akkkk Dec 22 '12 at 15:35
In the other question you like you have an equation: $x^2 = \frac{1}{x^2 + 1}$. Clearing the denominator, you have $x^4 + x^2 = 1$ which you just rewrite as $x^4 + x^2 - 1 = 0$... – Tyler Dec 22 '12 at 15:43

Essentially repeating Tyler Bailey's comment, in the answer to the linked question you faced solving $$x=\dfrac{1}{\sqrt{x^2+1}}.$$
Squaring both sides, noting this might introduce spurious solutions, gave $$x^2=\dfrac{1}{{x^2+1}}.$$
Multiplying both sides by $x^2+1$ (not numerator and denominator), which might lead to spurious solutions of the form $x=\pm i$ but in fact does not here, gave $$x^4 +x^2 =1.$$
Subtracting $1$ from both sides gave $$x^4 +x^2 -1 = 0.$$