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# If $23=x^4+\frac{1}{x^4}$, then what is the value of $x^2+\frac{1}{x^2}$?

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If $23=x^4+\frac{1}{x^4}$, then what is the value of $x^2+\frac{1}{x^2}$?

Apr 28, 2020

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Hi qwertyzz,

I'm really bad with these problems so this solution is probably wrong, but I'll give it a shot anyway!

If $$23=x^4+\frac{1}{x^4}$$ then what is the value of $$x^2+\frac{1}{x^2}$$?

So how I did this is:

First, I just set $$x^2+\frac{1}{x^2}=y$$

If you square both sides of $$x^2+\frac{1}{x^2}=y$$, you get $$(a^2)^2 + 2a^2\cdot\frac{1}{a^2} + \left(\frac{1}{a^2}\right)^{\!2} = y^2$$

So, this means that $$a^4 + \frac{1}{a^4} = y^2-2$$.

We know that $$a^4 + \frac{1}{a^4} =23$$, so $$y^2-2=23$$

So, we know $$\boxed{y= \pm5}$$

I hope this was right?

Please tell me if this is right!

:)

Apr 29, 2020
edited by lokiisnotdead  Apr 29, 2020
edited by lokiisnotdead  Apr 29, 2020
edited by lokiisnotdead  Apr 29, 2020
#2
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$$x^4+\frac{1}{x^4}=23$$  (1)

$$x^2+\frac{1}{x^2}$$ Square this

$$(x^2+\frac{1}{x^2})^2=x^4+\frac{1}{x^4}+2$$ From (1) we know that $$x^4+\frac{1}{x^4}=23$$ , substitute

$$(x^2+\frac{1}{x^2})^2=23+2=25$$

$$(x^2+\frac{1}{x^2})^2=25$$ Square root both sides

$$x^2+\frac{1}{x^2}=+5$$ or $$-5$$

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Apr 29, 2020