4
$\begingroup$

Sorry, I am not sure how to do the maths mark-up on this site but hopefully the question will make sense. I should know how to do this, but I have got myself stuck! Can anyone help?

$(x^2+x^{-2}-2)^{1/2}$

$\endgroup$
18
$\begingroup$

$$\left(x-\frac{1}{x}\right)^2= \dots?$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! Took me a moment to figure out what you were on about, but I do see how to simplify it now. So the smallest value I get factorising to get that and then substituting it in should be x-(1/x)... Assuming I understood you correctly, that is... $\endgroup$ – Magpie Jul 26 '12 at 16:55
  • $\begingroup$ @Magpie, I'm not sure what you mean by "the smalles value I get factorising...". The fact is your expression $\,x^2-2+x^{-2}\,$ has the expected form for the well-known squared binomial expression: (term 1 squared) + (term 2 squared) $\,\pm\,$ (twice term 1 times term 2) = (term 1 $\,\pm\,$ term 2)^2...practice, that's all. $\endgroup$ – DonAntonio Jul 26 '12 at 18:36
  • $\begingroup$ @DonAntonio Your answer is contradictory to jasoncube's. The expression is factored appropriately, however the initial exponent of $1/2$ should cancel out the exponent of $2$. Am I misunderstanding something? $\endgroup$ – user26649 Jul 27 '12 at 12:16
  • $\begingroup$ @FarhadYusufali, I think you are. I can't see how my answer is "contradictory to jasoncube's". IMO, both are accurate $\endgroup$ – DonAntonio Jul 27 '12 at 12:50
10
$\begingroup$

$$ \sqrt{x^2 + x^{-2} - 2} = \sqrt{x^2 – 2(x)(x^{-1})+ (x^{-1})^2} =\sqrt{(x – x^{-1})^2} = |x – x^{-1}| $$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

What is $x^{-2}$ equal to? Hint: can you write it as a fraction? If so, I would then look at adding and subtracting fractions and go from there.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

For $x\ne 0$, $x^2+x^{-2}-2=(x^4-2x^2+1)/x^2=(x^2-1)^2/x^2$. So taking square roots of both sides we get on the right side $|(x^2-1)/x|=|x-1/x|$.

| cite | improve this answer | |
$\endgroup$

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.