Real numbers

[willpower]

Let x be a real number such that: \[x^4 + \frac{1}{x^4}=2\] Find the possible values of \[x + \frac{1}{x}\].

[nafistiham]

step by step

\[x^{4}+ \frac {1} {x^{4}} = 2 \]
\[x^{8}-2x^{4}+1=0\]
\[(x^{4}-1)^{2}=0\]

\[x^{4}=1\]
\[x^{2}=1\]
\[x^{2}+ \frac {1} {x^{2}}=2\]

[willpower]

^How does that answer the question? We are required to find the possible values of \[x + \frac{1}{x}.\]

[nafistiham]

last step.

(as it is similar, wished that you would do it yourself. )

\[ x^{2}+\frac{1}{x^{2}}= 2\]
\[x^{4}-2x^{2}+1= 0\]
\[(x-1)^{2}= 0\]
\[x=\pm 1\]

so,firstly,

\[x+ \frac {1} {x}=2\]

or,

\[x+ \frac {1}{x}=-2\]

as a matter of fact you can see that,you can use any of the values you get through each of the step.