-3

Our teacher gave us a question but she didn't explain it. How to solve it?

$x^2+ \frac{1 }{x^2}=27$

Find, $x+ \frac{1}{x}$ and $x- \frac{1}{x}$

I am new to maths jax so sorry.

Rajdeep Singh
  • 73
  • 1
    The lack of parentheses makes it rather unclear what you're asking. Does $x^2+1/x^2$ mean $x^2 + \frac{1}{x^2}$, or does it mean $\frac{x^2+1}{x^2}$? Likewise with the other two expressions, which appear identical. – G Tony Jacobs Jun 13 '17 at 06:16
  • no he means the first – Saketh Malyala Jun 13 '17 at 06:16
  • Please use MathJax when formatting mathematics in this site. As it stands, I'm uncertain as to whether you mean $x^2+\frac1{x^2}=27$ or $\frac{x^2+1}{x^2}=27$. – Arthur Jun 13 '17 at 06:18
  • 4
    Literally every problem on this site is supposed to be "a mathematical problem," and there is no way to classify this question as "numerical lineral algebra." And "x+1/x and x+1/x" is odd. Can you spend some time trying to help us help you? – Thomas Andrews Jun 13 '17 at 06:18
  • The first one. (x²) +1/x² – Rajdeep Singh Jun 13 '17 at 06:18
  • 2
    What's $(x+1/x)^2$? – Angina Seng Jun 13 '17 at 06:21
  • I think you need to improve the question in multiple ways: (1) Choose a better title (2) use MathJAX (3) select your tags properly (4) proof read the question (why do you write find $x+1/x$ and $x+1/x$? those are the same). – skyking Jun 13 '17 at 06:35
  • @RajdeepSingh I edited the question according to what I understood from the comments and the answer. – Majid Jun 13 '17 at 06:39

1 Answers1

1

Alright this is a fun one.

If $\displaystyle x^2 + \frac{1}{x^2}=27$, then $\displaystyle x^2 - 2 + \frac{1}{x^2} = 25$.

Because $\displaystyle x^2-2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2$, we have $\displaystyle \boxed{x-\frac{1}{x}=\pm5}$

If $\displaystyle x^2 + \frac{1}{x^2}=27$, then $\displaystyle x^2 + 2 + \frac{1}{x^2} = 29$.

Because $\displaystyle x^2+2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2$, we have $\displaystyle \boxed{x+\frac{1}{x}=\pm \sqrt{29}}$

*** Notice that $\displaystyle \left(x \pm \frac{1}{x}\right)^2=x^2 \pm x\left(\frac1x\right)\pm x\left(\frac1x\right)+\frac{1}{x^2}=x^2\pm2+\frac{1}{x^2}$

Saketh Malyala
  • 13,637