9.6 Solve equations with square roots  (Page 2/6)

Solution

	$\sqrt{r+4}-r+2=0$
Isolate the radical.	$\sqrt{r+4}=r-2$
Square both sides of the equation.	${(\sqrt{r+4})}^2={\left(r-2\right)}^2$
Solve the new equation.	$r+4=r^2-4r+4$
It is a quadratic equation, so get zero on one side.	$0=r^2-5r$
Factor the right side.	$0=r(r-5)$
Use the zero product property.	$0=r0=r-5$
Solve the equation.	$r=0r=5$
Check the answer.
	The solution is $r=5$ .
	$r=0$ is an extraneous solution.

Solution

	$3\sqrt{3x-5}-8=4$
Isolate the radical.	$3\sqrt{3x-5}=12$
Square both sides of the equation.	${\left(3\sqrt{3x-5}\right)}^2={\left(12\right)}^2$
Simplify, then solve the new equation.	$9(3x-5)=144$
Distribute.	$27x-45=144$
Solve the equation.	$27x=189$
	$x=7$
Check the answer.
	The solution is $x=7$ .

Solution

$\begin{array}{cccc}& & & \sqrt{4z-3}=\sqrt{3z+2}\hfill \\ \\ \\ \text{The radical terms are isolated.}\hfill & & & \sqrt{4z-3}=\sqrt{3z+2}\hfill \\ \\ \\ \text{Square both sides of the equation.}\hfill & & & {\left(\sqrt{4z-3}\right)}^2={\left(\sqrt{3z+2}\right)}^2\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & \begin{array}{c}4z-3=3z+2\hfill \\ z-3=2\hfill \\ z=5\hfill \end{array}\hfill \\ \\ \\ \text{Check the answer.}\hfill & & & \\ \text{We leave it to you to show that 5 checks!}\hfill & & & \text{The solution is}z=5.\hfill \end{array}$

Solution

$\begin{array}{cccc}& & & \sqrt{m}+1=\sqrt{m+9}\hfill \\ \\ \\ \text{The radical on the right side is isolated. Square both sides.}\hfill & & & {\left(\sqrt{m}+1\right)}^2={\left(\sqrt{m+9}\right)}^2\hfill \\ \\ \\ \text{Simplify—be very careful as you multiply!}\hfill & & & m+2\sqrt{m}+1=m+9\hfill \\ \text{There is still a radical in the equation.}\hfill & & & \\ \text{So we must repeat the previous steps. Isolate the radical.}\hfill & & & 2\sqrt{m}=8\hfill \\ \\ \\ \text{Square both sides.}\hfill & & & {\left(2\sqrt{m}\right)}^2={\left(8\right)}^2\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & 4m=64\hfill \\ & & & m=16\hfill \\ \\ \\ \text{Check the answer.}\hfill & & & \\ \\ \\ \text{We leave it to you to show that}m=16\text{checks!}\hfill & & & \text{The solution is}m=16.\hfill \end{array}$

Solution

$\begin{array}{cccc}& & & \sqrt{q-2}+3=\sqrt{4q+1}\hfill \\ \\ \\ \begin{array}{c}\text{The radical on the right side is isolated.}\hfill \\ \text{Square both sides.}\hfill \end{array}\hfill & & & {\left(\sqrt{q-2}+3\right)}^2={\left(\sqrt{4q+1}\right)}^2\hfill \\ \\ \\ \text{Simplify.}\hfill & & & q-2+6\sqrt{q-2}+9=4q+1\hfill \\ \\ \\ \begin{array}{c}\text{There is still a radical in the equation. So}\hfill \\ \text{we must repeat the previous steps. Isolate}\hfill \\ \text{the radical.}\hfill \end{array}\hfill & & & 6\sqrt{q-2}=3q-6\hfill \\ \\ \\ \text{Square both sides.}\hfill & & & {\left(6\sqrt{q-2}\right)}^2={\left(3q-6\right)}^2\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & 36\left(q-2\right)=9q^2-36q+36\hfill \\ \\ \\ \text{Distribute.}\hfill & & & 36q-72=9q^2-36q+36\hfill \\ \\ \\ \begin{array}{c}\text{It is a quadratic equation, so get zero on}\hfill \\ \text{one side.}\hfill \end{array}\hfill & & & 0=9q^2-72q+108\hfill \\ \\ \\ \text{Factor the right side.}\hfill & & & \begin{array}{c}0=9\left(q^2-8q+12\right)\hfill \\ 0=9\left(q-6\right)\left(q-2\right)\hfill \end{array}\hfill \\ \\ \\ \text{Use the zero product property.}\hfill & & & \begin{array}{cccc}q-6=0\hfill & & & q-2=0\hfill \\ q=6\hfill & & & q=2\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{The checks are left to you. (Both solutions}\hfill \\ \text{should work.)}\hfill \end{array}\hfill & & & \text{The solutions are}q=6\text{and}q=2.\hfill \end{array}$