Solve Equations with Square Roots: example 9.80

Solve: $\sqrt{r + 4} - r + 2 = 0$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve: $\sqrt{m + 9} - m + 3 = 0$ .

How did it go?

	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	${(\sqrt{r + 4})}^{2} = {(r - 2)}^{2}$
Solve the new equation.	$r + 4 = r^{2} - 4 r + 4$
It is a quadratic equation, so get zero on one side.	$0 = r^{2} - 5 r$
Factor the right side.	$0 = r (r - 5)$
Use the zero product property.	$0 = r 0 = r - 5$
Solve the equation.	$r = 0 r = 5$
Check the answer.
	The solution is $r = 5$ .
	$r = 0$ is an extraneous solution.

	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	${(\sqrt{r + 4})}^{2} = {(r - 2)}^{2}$
Solve the new equation.	$r + 4 = r^{2} - 4 r + 4$
It is a quadratic equation, so get zero on one side.	$0 = r^{2} - 5 r$
Factor the right side.	$0 = r (r - 5)$
Use the zero product property.	$0 = r 0 = r - 5$
Solve the equation.	$r = 0 r = 5$
Check the answer.
	The solution is $r = 5$ .
	$r = 0$ is an extraneous solution.