4

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?

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

Myshkin
  • 35,974
  • 27
  • 154
  • 332
Magpie
  • 523

4 Answers4

18

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

Michael Hardy
  • 1
DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • 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... – Magpie Jul 26 '12 at 16:55
  • @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. – DonAntonio Jul 26 '12 at 18:36
  • @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? –  Jul 27 '12 at 12:16
  • @FarhadYusufali, I think you are. I can't see how my answer is "contradictory to jasoncube's". IMO, both are accurate – DonAntonio Jul 27 '12 at 12:50
10

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

Norbert
  • 56,803
JSCB
  • 13,456
  • 15
  • 59
  • 123
2

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.

Code-Guru
  • 2,176
1

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 $|(x^2-1)/x|=|x-1/x|$.

i. m. soloveichik
  • 4,259