Math help.

**devi** · August 21st 2008, 10:28 PM

If $(x-1/x) = 5 the value of x^2+1/x is$
(a)23
(b)27
(c)25
(d)29

**wisterville** · August 22nd 2008, 05:49 AM

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.

**Soroban** · August 22nd 2008, 02:39 PM

Hello, devi!

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

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

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