Chapter 12.5, Problem 31WE

To determine

Solve the quadratic equation.

Answer to Problem 31WE

  $x\in \left\{-\frac{2\sqrt{6}}{3},\frac{2\sqrt{6}}{3}\right\}$

Explanation of Solution

Given:

The equation is :

  $9x^4-64=0$

Calculation:

The given equation $9x^4-64=0\Rightarrow 9{(x^2)}^2-64=0$ .

Substitute $x^2=y$

So,

  $\begin{array}{l}9y^2-64=0\\ \Rightarrow 9y^2=\mathrm{64.....................................}[\text{add 64 each side]}\\ \Rightarrow y^2=\frac{64}{9}\mathrm{......................................}[\text{divide each side by 9]}\\ \Rightarrow y=\pm \sqrt{\frac{64}{9}}\mathrm{...................................}[\text{take square root each side]}\\ \Rightarrow y=\frac{8}{3}\text{ or y}=-\frac{8}{3}\\ \Rightarrow x^2=\frac{8}{3}\text{ or }{x}^2=-\frac{8}{3}\mathrm{.........................}[y=x^2]\\ \Rightarrow x=\pm \sqrt{\frac{8}{3}}\cdot \sqrt{\frac{3}{3}}\text{ or x}=\pm \sqrt{-\frac{8}{3}}\\ \Rightarrow x=\pm \frac{2\sqrt{6}}{3}\end{array}$

Since square root take only non-negative values, so the second equation has no real roots.

Hence , $x\in \left\{-\frac{2\sqrt{6}}{3},\frac{2\sqrt{6}}{3}\right\}$