Question

Solve the quadratic equation.

$x^4-14x^2+24=0$

Short answer

The solutions are$x=\pm \sqrt{2}$and$x=\pm 2\sqrt{3}$.

Step-by-step solution

Step 1. Given information.

The given equation is$x^4-14x^2+24=0$.

Step 2. Rewrite the equation by substituting x2=u.

$\begin{array}{rcl}{x}^4-14x^2+24& =& 0\\ {\left(x^2\right)}^2-14x^2+24& =& 0\\ u^2-14u+24& =& 0\cdots \cdots \cdots \left(x^2=u\right)\end{array}$

Step 3. Simplify the above equation by factorization.

$\begin{array}{rcl}{u}^2-14u+24& =& 0\\ u^2-12u-2u+24& =& 0\\ u\left(u-12\right)-2\left(u-12\right)& =& 0\\ \left(u-12\right)\left(u-2\right)& =& 0\end{array}$

Step 4. Now apply the zero product property and solve for u.

$\begin{array}{rcl}u-12& =& 0\\ u& =& 12\end{array}$and$\begin{array}{rcl}u-2& =& 0\\ u& =& 2\end{array}$

Step 5. Replace u with x2 and solve for x.

$\begin{array}{rcl}u& =& 12\\ x^2& =& 12\\ x& =& \pm \sqrt{12}\\ x& =& \pm 2\sqrt{3}\end{array}$ and$\begin{array}{rcl}u& =& 2\\ x^2& =& 2\\ x& =& \pm \sqrt{2}\end{array}$

Step 6. Simplified answer.

Hence, the required solutions are $x=\pm \sqrt{2}$and$x=\pm 2\sqrt{3}$.