Question

Solve the quadratic equation.

$4x^4-5x^2+1=0$

Short answer

The solutions are$x=\pm \frac{1}{2}$and$x=\pm 1$.

Step-by-step solution

Step 1. Given information.

The given equation is$4x^4-5x^2+1=0$.

Step 2. Simplify the equation by the factorization method.

$\begin{array}{rcl}4x^4-5x^2+1& =& 0\\ 4{\left(x^2\right)}^2-5x^2+1& =& 0\\ 4{\left(x^2\right)}^2-x^2-4x^2+1& =& 0\\ x^2\left(4x^2-1\right)-1\left(4x^2-1\right)& =& 0\\ \left(4x^2-1\right)\left(x^2-1\right)& =& 0\end{array}$

Step 3. Solve the equation for x.

$\begin{array}{rcl}4x^2-1& =& 0\\ 4x^2& =& 1\\ x^2& =& \frac{1}{4}\\ x& =& \pm \sqrt{\frac{1}{4}}\\ x& =& \pm \frac{1}{2}\end{array}$ and $\begin{array}{rcl}{x}^2-1& =& 0\\ x^2& =& 1\\ x& =& \pm \sqrt{1}\\ x& =& \pm 1\end{array}$

Step 4. Simplified answer.

Hence, the required solutions are$x=\pm \frac{1}{2}$and$x=\pm 1$.