## Intermediate Algebra (6th Edition)

The solutions are $-\sqrt{11}, -1, 1, \text{ and } \sqrt{11}$.
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 $-\sqrt{11}, -1, 1, \text{ and } \sqrt{11}$.