Chapter 12, Problem 1CR

To determine

To find: The solutions real and complex of the equation.

Answer to Problem 1CR

The solution set of the equation is $\left\{-3,3,-3i,3i\right\}$ .

Explanation of Solution

Given information:

The given equation is $\left|x^2\right|=9$ .

Calculation:

The equation $\left|x^2\right|=9$ is of the form $\left|u\right|=a$

If $a$ is a positive real number and if u is any algebraic expression, them $\left|u\right|=a$ is equivalent to $u=a$ or $u=-a$ .

Thus, $\left|x^2\right|=9$ is equivalent to $x^2=9$ or $x^2=-9$ .

According to the Square Root Method, if $x^2=p$ and $p\ge 0$ , then $x=\sqrt{p}$ or $x=-\sqrt{p}$ .

  $\begin{array}{c}{x}^2=9\\ x=\pm \sqrt{9}\\ x=3or\\ x=-3\end{array}$

Next, consider $x^2=-9$ .

Use the Square Root Method.

  $x=\pm \sqrt{-9}$

The principal square root of $-N,\sqrt{-N}=\sqrt{N}i$ where N is any positive real number.

So,

  $\sqrt{-9}=\sqrt{9}i=\pm 3i$

Thus, $x=\pm 3i$ .

The equation has four solutions $3,-3,-3i$ and $3i$ .

Therefore, the solution set of the equation is $\left\{-3,3,-3i,3i\right\}$ .