Chapter 1.2, Problem 2QR

Solution

To solve: The given equation or inequality.

  $x=\left\{\pm 3\right\}$

Given Information:

The equation $9-x^2=0$ .

Concept used:

Use the formula:

  $a^2-x^2=\left(a+x\right)\left(a-x\right)$ .

Calculation:

Use the formula and simplify the equation as follows,

  $\begin{array}{c}9-x^2=0\\ 3^2-x^2=0\\ \left(3-x\right)\left(3+x\right)=0\end{array}$

Since, when the product two terms is zero then one of them must be zero.

Hence,

  $\left(3-x\right)=0\text{or}\left(3+x\right)=0$

Set the first factors equal to zero and simplify for the value of $x$ as follows.

  $\begin{array}{c}3-x=0\\ x=3\end{array}$

Set the first factors equal to zero and simplify for the value of $x$ as follows.

  $\begin{array}{c}3+x=0\\ 3=-x\\ x=-3\end{array}$

Hence, $x=3\text{or}\text{-3}$

Therefore, the solution set is $x=\left\{\pm 3\right\}$ .