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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
 #1
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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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+1

\(x^4+\frac{1}{x^4}=23\)  (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 \(-5\)

.
 Apr 29, 2020

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