How do you solve $x^4 + x^2 = 1$ ?

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Answer 1 · 2016-03-14T20:24:07

Solve as a quadratic in $x^2$ using the quadratic formula, then take square roots...

$x^4+x^2=1$

Subtract $1$ from both sides to get:

$x^4+x^2-1 = 0$

Writing $x^4 = (x^2)^2$ we have:

$(x^2)^2+(x^2)-1 = 0$

This is in the form $aX^2+bX+c = 0$ with $X=x^2$ , $a=1$ , $b=1$ and $c=-1$ .

We can use the quadratic formula to find:

$x^2 = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-1+-sqrt(1^2-(4*1*-1)))/(2*1)$

$=(-1+-sqrt(5))/2$

So:

$x = +-sqrt((-1+sqrt(5))/2)$

Or:

$x = +-sqrt((-1-sqrt(5))/2) = +-sqrt((1+sqrt(5))/2)i$

How do you solve x^4 + x^2 = 1 x^4 + x^2 = 1 ?