Chapter 5.6, Problem 50SR

To determine

To solve: the equation $4x^4-8x^2-96=0$ .

Answer to Problem 50SR

The solutions are $\sqrt{6},-\sqrt{6},2i,\text{ and }-2i$ .

Explanation of Solution

Given information:

  $4x^4-8x^2-96=0$ .

Formula used:

An expression that is quadratic form can be written as $a{u}^2+bu+c$ for any numbers a , b , and c , $a\ne 0$ , where u is some expression in x . The expression $a{u}^2+bu+c$ is called the quadratic form of the original expression.

Calculation:

Consider ,

  $4x^4-8x^2-96=0$

  $={\left(2x^2\right)}^2-4\left(2x^2\right)-96=0$

Let $2x^2=u$ .

  $\begin{array}{l}=u^2-4u-96=0\\ =u^2-12u+8u-96=0\\ =u(u-12)+8(u-12)=0\\ =(u+8)(u-12)=0\end{array}$

  $\begin{array}{l}u+8=0\\ u=-8\\ 2x^2=-8\\ \frac{2x^2}{2}=\frac{-8}{2}\\ x^2=-4\\ x=\pm \sqrt{4i^2}\\ x=\pm 2i\end{array}$ or $\begin{array}{l}u-12=0\\ u=12\\ 2x^2=12\\ \frac{2x^2}{2}=\frac{12}{2}\\ x^2=6\\ x=\pm \sqrt{6}\end{array}$

Therefore, the solutions of the equations are $\sqrt{6},-\sqrt{6},2i,\text{ and }-2i$