Chapter 6.5, Problem 58PPS

To determine

To solve: the given equation.

Answer to Problem 58PPS

  $x=\pm \frac{1}{2},\pm \sqrt{\frac{3}{2}}$ .

Explanation of Solution

Given:

The given equation is $8x^4+10x^2-3=0$ .

Calculation:

The given equation is $8x^4+10x^2-3=0$ .

In order to solve the equation, factor the equation. Therefore,

  $\begin{array}{l}8x^4+10x^2-3=0\left\{\text{given equation}\right\}\\ 8x^4-2x^2+12x^2-3=0\left\{\text{express in simplified form}\right\}\\ 2x^2\left(4x^2-1\right)-3\left(4x^2-1\right)=0\left\{\text{taking }\left(4x^2-1\right)\text{as common}\right\}\\ \left(4x^2-1\right)\left(2x^2-3\right)=0\left\{\text{use distributive property of product}\right\}\end{array}$

Either $\left(2x^2-3\right)=0$ or $\left(4x^2-1\right)=0$

when $\left(2x^2-3\right)=0$ , then $2x^2=3\Rightarrow x^2=\frac{3}{2}\Rightarrow x=\pm \sqrt{\frac{3}{2}}$

when $\left(4x^2-1\right)=0$ , then $4x^2=1;x^2=\frac{1}{4}\Rightarrow x=\pm \sqrt{\frac{1}{4}}=\pm \frac{1}{2}$ :

Hence, the solution of the given equation is $x=\pm \frac{1}{2},\pm \sqrt{\frac{3}{2}}$ .