# Math help.

• August 21st 2008, 10:28 PM
devi
Math help.
If $(x-1/x) = 5 the value of x^2+1/x is$
(a)23
(b)27
(c)25
(d)29
• August 22nd 2008, 05:49 AM
wisterville
Hello,

I think something is wrong...
If you are asked the value of $x^2+\frac{1}{x^2}$, compute $\left(x-\frac{1}{x}\right)^2$ and compare.

If you really want $x^2+\frac{1}{x}$, the easiest way may be to compute the value of x by $x^2-1=5x$.

Bye.
• August 22nd 2008, 02:39 PM
Soroban
Hello, devi!

You really should learn to control the text when using LaTeX.
$Otherwise everything looks like this.$

Quote:

If $x-\frac{1}{x} \:=\:5$, the value of $x^2+\frac{1}{x^2}$ is:

. . $(a)\;23 \qquad(b)\;27 \qquad (c)\;25 \qquad(d)\;29$

We are given: . $x - \frac{1}{x} \:=\:5$

Square both sides: . $\left(x-\frac{1}{x}\right)^2 \:=\:5^2 \quad\Rightarrow\quad x^2 - 2 + \frac{1}{x^2} \;=\;25$

Therefore: . $x^2 + \frac{1}{x^2} \;=\;27\;\;(b)$