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Solve: $\sqrt{r+4}-r+2=0$ .
$\sqrt{r+4}-r+2=0$ | |
Isolate the radical. | $\phantom{\rule{3.17em}{0ex}}\sqrt{r+4}=r-2$ |
Square both sides of the equation. | $\phantom{\rule{2.26em}{0ex}}{(\sqrt{r+4})}^{2}={\left(r-2\right)}^{2}$ |
Solve the new equation. | $\phantom{\rule{3.63em}{0ex}}r+4={r}^{2}-4r+4$ |
It is a quadratic equation, so get zero on one side. | $\phantom{\rule{5.17em}{0ex}}0={r}^{2}-5r$ |
Factor the right side. | $\phantom{\rule{5.17em}{0ex}}0=r(r-5)$ |
Use the zero product property. | $\phantom{\rule{1.99em}{0ex}}0=r\phantom{\rule{1em}{0ex}}0=r-5$ |
Solve the equation. | $\phantom{\rule{2.09em}{0ex}}r=0\phantom{\rule{1em}{0ex}}r=5$ |
Check the answer. | |
The solution is $r=5$ . | |
$r=0$ is an extraneous solution. |
When there is a coefficient in front of the radical, we must square it, too.
Solve: $3\sqrt{3x-5}-8=4$ .
$3\sqrt{3x-5}-8=4$ | |
Isolate the radical. | $\phantom{\rule{1.62em}{0ex}}3\sqrt{3x-5}=12$ |
Square both sides of the equation. | $\phantom{\rule{0.67em}{0ex}}{\left(3\sqrt{3x-5}\right)}^{2}={\left(12\right)}^{2}$ |
Simplify, then solve the new equation. | $\phantom{\rule{1.47em}{0ex}}9(3x-5)=144$ |
Distribute. | $\phantom{\rule{1.59em}{0ex}}27x-45=144$ |
Solve the equation. | $\phantom{\rule{3.7em}{0ex}}27x=189$ |
$\phantom{\rule{4.76em}{0ex}}x=7$ | |
Check the answer. | |
The solution is $x=7$ . |
Solve: $\sqrt{4z-3}=\sqrt{3z+2}$ .
$\begin{array}{cccc}& & & \phantom{\rule{5em}{0ex}}\sqrt{4z-3}=\sqrt{3z+2}\hfill \\ \\ \\ \text{The radical terms are isolated.}\hfill & & & \phantom{\rule{5em}{0ex}}\sqrt{4z-3}=\sqrt{3z+2}\hfill \\ \\ \\ \text{Square both sides of the equation.}\hfill & & & \phantom{\rule{4.07em}{0ex}}{\left(\sqrt{4z-3}\right)}^{2}={\left(\sqrt{3z+2}\right)}^{2}\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & \begin{array}{c}\phantom{\rule{5.5em}{0ex}}4z-3=3z+2\hfill \\ \phantom{\rule{6em}{0ex}}z-3=2\hfill \\ \phantom{\rule{7.61em}{0ex}}z=5\hfill \end{array}\hfill \\ \\ \\ \text{Check the answer.}\hfill & & & \\ \text{We leave it to you to show that 5 checks!}\hfill & & & \phantom{\rule{4em}{0ex}}\text{The solution is}\phantom{\rule{0.2em}{0ex}}z=5.\hfill \end{array}$
Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.
Solve: $\sqrt{m}+1=\sqrt{m+9}$ .
$\begin{array}{cccc}& & & \phantom{\rule{2.43em}{0ex}}\sqrt{m}+1=\sqrt{m+9}\hfill \\ \\ \\ \text{The radical on the right side is isolated. Square both sides.}\hfill & & & \phantom{\rule{1.3em}{0ex}}{\left(\sqrt{m}+1\right)}^{2}={\left(\sqrt{m+9}\right)}^{2}\hfill \\ \\ \\ \text{Simplify\u2014be very careful as you multiply!}\hfill & & & \phantom{\rule{0.09em}{0ex}}m+2\sqrt{m}+1=m+9\hfill \\ \text{There is still a radical in the equation.}\hfill & & & \\ \text{So we must repeat the previous steps. Isolate the radical.}\hfill & & & \phantom{\rule{3.58em}{0ex}}2\sqrt{m}=8\hfill \\ \\ \\ \text{Square both sides.}\hfill & & & \phantom{\rule{2.43em}{0ex}}{\left(2\sqrt{m}\right)}^{2}={\left(8\right)}^{2}\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & \phantom{\rule{4.04em}{0ex}}4m=64\hfill \\ & & & \phantom{\rule{4.54em}{0ex}}m=16\hfill \\ \\ \\ \text{Check the answer.}\hfill & & & \\ \\ \\ \text{We leave it to you to show that}\phantom{\rule{0.2em}{0ex}}m=16\phantom{\rule{0.2em}{0ex}}\text{checks!}\hfill & & & \text{The solution is}\phantom{\rule{0.2em}{0ex}}m=16.\hfill \end{array}$
Solve: $\sqrt{q-2}+3=\sqrt{4q+1}$ .
$\begin{array}{cccc}& & & \phantom{\rule{7.29em}{0ex}}\sqrt{q-2}+3=\sqrt{4q+1}\hfill \\ \\ \\ \begin{array}{c}\text{The radical on the right side is isolated.}\hfill \\ \text{Square both sides.}\hfill \end{array}\hfill & & & \phantom{\rule{6.34em}{0ex}}{\left(\sqrt{q-2}+3\right)}^{2}={\left(\sqrt{4q+1}\right)}^{2}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{3.57em}{0ex}}q-2+6\sqrt{q-2}+9=4q+1\hfill \\ \\ \\ \begin{array}{c}\text{There is still a radical in the equation. So}\hfill \\ \text{we must repeat the previous steps. Isolate}\hfill \\ \text{the radical.}\hfill \end{array}\hfill & & & \phantom{\rule{8.41em}{0ex}}6\sqrt{q-2}=3q-6\hfill \\ \\ \\ \text{Square both sides.}\hfill & & & \phantom{\rule{7.52em}{0ex}}{\left(6\sqrt{q-2}\right)}^{2}={\left(3q-6\right)}^{2}\hfill \\ \\ \\ \text{Simplify, then solve the new equation.}\hfill & & & \phantom{\rule{8.11em}{0ex}}36\left(q-2\right)=9{q}^{2}-36q+36\hfill \\ \\ \\ \text{Distribute.}\hfill & & & \phantom{\rule{7.88em}{0ex}}36q-72=9{q}^{2}-36q+36\hfill \\ \\ \\ \begin{array}{c}\text{It is a quadratic equation, so get zero on}\hfill \\ \text{one side.}\hfill \end{array}\hfill & & & \phantom{\rule{11.06em}{0ex}}0=9{q}^{2}-72q+108\hfill \\ \\ \\ \text{Factor the right side.}\hfill & & & \begin{array}{c}\phantom{\rule{11.06em}{0ex}}0=9\left({q}^{2}-8q+12\right)\hfill \\ \phantom{\rule{11.06em}{0ex}}0=9\left(q-6\right)\left(q-2\right)\hfill \end{array}\hfill \\ \\ \\ \text{Use the zero product property.}\hfill & & & \begin{array}{cccc}\phantom{\rule{4.05em}{0ex}}q-6=0\hfill & & & q-2=0\hfill \\ \phantom{\rule{5.65em}{0ex}}q=6\hfill & & & \phantom{\rule{1.65em}{0ex}}q=2\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{The checks are left to you. (Both solutions}\hfill \\ \text{should work.)}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\text{The solutions are}\phantom{\rule{0.2em}{0ex}}q=6\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}q=2.\hfill \end{array}$
As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.
We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.
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