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

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

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Answer 1 · 2016-03-14T20:24:07

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

Explanation:

$x^{4} + x^{2} = 1$ $x^4+x^2=1$

Subtract $1$ $1$ from both sides to get:

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

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

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

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

We can use the quadratic formula to find:

$x^{2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x^2 = (-b+-sqrt(b^2-4ac))/(2a)$

$= \frac{- 1 \pm \sqrt{1^{2} - (4 \cdot 1 \cdot - 1)}}{2 \cdot 1}$ $=(-1+-sqrt(1^2-(4*1*-1)))/(2*1)$

$= \frac{- 1 \pm \sqrt{5}}{2}$ $=(-1+-sqrt(5))/2$

So:

$x = \pm \sqrt{\frac{- 1 + \sqrt{5}}{2}}$ $x = +-sqrt((-1+sqrt(5))/2)$

Or:

$x = \pm \sqrt{\frac{- 1 - \sqrt{5}}{2}} = \pm \sqrt{\frac{1 + \sqrt{5}}{2}} i$ $x = +-sqrt((-1-sqrt(5))/2) = +-sqrt((1+sqrt(5))/2)i$

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

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