Question

Solve the quadratic equation.

$x^4+4x^2-32=0$

Short answer

The solutions are$x=\pm 2$and$x=\pm 2\sqrt{2}i$.

Step-by-step solution

Step 1. Given information.

The given equation is$x^4+4x^2-32=0$.

Step 2. Simplify the equation by factorization.

$\begin{array}{rcl}{x}^4+4x^2-32& =& 0\\ {\left(x^2\right)}^2+4x^2-32& =& 0\\ {\left(x^2\right)}^2+8x^2-4x^2-32& =& 0\\ x^2\left(x^2+8\right)-4\left(x^2+8\right)& =& 0\\ \left(x^2+8\right)\left(x^2-4\right)& =& 0\end{array}$

Step 3. Apply the zero product property and solve for x.

$\begin{array}{rcl}{x}^2+8& =& 0\\ x^2& =& -8\\ x& =& \pm \sqrt{-8}\\ & =& \pm \sqrt{8}\sqrt{-1}\\ & =& \pm 2\sqrt{2}i\cdots \cdots \cdots \left(i=\sqrt{-1}\right)\end{array}$ and $\begin{array}{rcl}{x}^2-4& =& 0\\ x^2& =& 4\\ x& =& \pm \sqrt{4}\\ & =& \pm 2\end{array}$

Step 4. Simplified answer.

Hence, the required solutions are$x=\pm 2$and$x=\pm 2\sqrt{2}i$.