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I am not sure how to approach this question:

You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.

I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not.

I need help in getting started.

mikoyan
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3 Answers3

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Notice $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.

Yiyuan Lee
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Observe that
$$x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right )^2-2.$$
Hope this helps. :)

MJD
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awllower
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Thank you to Yiyuan Lee and awllower who really answered the question.

I am only adding this for completeness, and to remind myself of the solution.

We are given $x + \frac{1}{x} =7$ and want to use this to find $x^2 + \frac{1}{x^2}$.

Using the fact $$\left(x +\frac{1}{x}\right)^2 = x^2 +\frac{1}{x^2} +2$$ and rewriting this as: $$\left(x +\frac{1}{x}\right)^2 -2 = x^2 +\frac{1}{x^2}$$ We can substitute $x + \frac{1}{x} =7$ to get: $$(7)^2 -2 = x^2 +\frac{1}{x^2}$$ So $$x^2 +\frac{1}{x^2} = 47$$ The same method can be used to calculate $x^3 + \frac{1}{x^3}$: $$\left(x +\frac{1}{x}\right)^3 = x^3 +\frac{1}{x^3} +3x +\frac{3}{x}$$ Noticing that $3x +\frac{3}{x}$ = $3(x+ \frac{1}{x})$, which means $3x +\frac{3}{x} = 3.(7) = 21$

Then it is possible to write: $$\left(x +\frac{1}{x}\right)^3 -21 = x^3 + \frac{1}{x^3}$$ $$(7)^3 -21 = x^3 + \frac{1}{x^3}$$ $$x^3 + \frac{1}{x^3} = 343 -21$$ $$x^3 + \frac{1}{x^3} = 322$$

mikoyan
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  • In fact, the same approach can be used to find $x^n + \frac{1}{x^n}$ for $4,5,6...$. – Yiyuan Lee Feb 17 '14 at 22:13
  • I was thinking of ways to extend the approach to show it for the nth case. Thank you again for your help and interest. – mikoyan Feb 18 '14 at 09:11
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    Use $(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})=x^5+\frac{1}{x^5}+(x+\frac{1}{x})$ to find $x^5+\frac{1}{x^5}$ and generalize it to other powers. – rah4927 Feb 19 '14 at 08:05
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    Or, you could generalize it with a recursive general formula, perhaps through induction. I would find such a formula more useful. – Yiyuan Lee Feb 19 '14 at 14:29
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    @mikoyan Since it seems you are interested, I have a challenge for you: If given that $x+\frac{1}{x}= \sqrt{3}$, then find, $$x^{204}+\frac{1}{x^{204}}$$ – Sawarnik Feb 23 '14 at 08:07