Answer
The solutions are $-\sqrt{11}, -1, 1, \text{ and } \sqrt{11}$.
Work Step by Step
The given equation can be written as:
$(x^2)^2-12x^2+11=0$
Let $u=x^2$.
Rewrite the equation above using $u$ to obtain:
$u^2-12u+11=0$
Factor the trinomial to obtain:
$(u-11)(u-1)=0$
Factor each factor with a variable to zero then solve each equation to obtain:
\begin{array}{ccc}
&u-11=0 &\text{or} &u-1=0
\\&u=11 &\text{or} &u=1
\end{array}
Since $u=x^2$, then:
\begin{array}{ccc}
&u=11&\text{or} &u=1
\\&x^2=11 &\text{or} &x^2=1
\\&x=\pm\sqrt{11} &\text{or} &u=\pm \sqrt1
\\&x=\pm\sqrt{11} &\text{or} &u=\pm 1
\end{array}
Therefore, the solutions are $-\sqrt{11}, -1, 1, \text{ and } \sqrt{11}$.
