Answer
$$x = \pm \sqrt {\frac{{ - 1 + \sqrt 5 }}{2}} $$
Work Step by Step
$$\eqalign{
& {x^4} + {x^2} - 1 = 0 \cr
& {\text{Rewrite}} \cr
& {\left( {{x^2}} \right)^2} + \left( {{x^2}} \right) - 1 = 0 \cr
& {\text{Let }}u = {x^2} \cr
& {u^2} + u - 1 = 0 \cr
& {\text{By the quadratic formula}} \cr
& u = \frac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} \cr
& u = \frac{{ - 1 \pm \sqrt 5 }}{2} \cr
& u = \frac{{ - 1 + \sqrt 5 }}{2},{\text{ }}u = \frac{{ - 1 - \sqrt 5 }}{2} \cr
& {\text{Write in terms of }}x \cr
& {x^2} = \frac{{ - 1 + \sqrt 5 }}{2},{\text{ }}\underbrace {{x^2} = \frac{{ - 1 - \sqrt 5 }}{2}}_{{\text{No real solutions}}} \cr
& x = \pm \sqrt {\frac{{ - 1 + \sqrt 5 }}{2}} \cr} $$
