10.1 Solve quadratic equations using the square root property  (Page 2/5)

Solution

The quadratic term is isolated.	$5m^2=80$
Divide by 5 to make its cofficient 1.	$\frac{5m^2}{5}=\frac{80}{5}$
Simplify.	$m^2=16$
Use the Square Root Property.	$m=\pm \sqrt{16}$
Simplify the radical.	$m=\pm 4$
Rewrite to show two solutions.	$m=4,m=\text{−}4$
Check the solutions.

Solution

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \text{Isolate the quadratic term.}\hfill \\ \text{Use the Square Root Property.}\hfill \end{array}\hfill & & & \hfill \begin{array}{ccc}\hfill q^2+24& =\hfill & 0\hfill \\ \hfill q^2& =\hfill & \mathrm{-24}\hfill \\ \hfill q& =\hfill & \pm \sqrt{\mathrm{-24}}\hfill \end{array}\\ \text{The}\sqrt{\mathrm{-24}}\text{is not a real number.}\hfill & & & \hfill \text{There is no real solution.}\hfill \end{array}$

Solution

	$\frac{2}{3}{u}^2+5=17$
Isolate the quadratic term.	$\frac{2}{3}{u}^2=12$
Multiply by $\frac{3}{2}$ to make the coefficient 1.	$\frac{3}{2}·\frac{2}{3}{u}^2=\frac{3}{2}·12$
Simplify.	$u^2=18$
Use the Square Root Property.	$u=\pm \sqrt{18}$
Simplify the radical.	$u=\pm \sqrt{9}\sqrt{2}$
Simplify.	$u=\pm 3\sqrt{2}$
Rewrite to show two solutions.	$u=3\sqrt{2},u=\text{−}3\sqrt{2}$
Check.

Solution

$\begin{array}{c}\begin{array}{cccccc}& & & \hfill 2c^2-4& =\hfill & 45\hfill \\ \text{Isolate the quadratic term.}\hfill & & & \hfill 2c^2& =\hfill & 49\hfill \\ \text{Divide by 2 to make the coefficient 1.}\hfill & & & \hfill \frac{2c^2}{2}& =\hfill & \frac{49}{2}\hfill \\ \text{Simplify.}\hfill & & & \hfill c^2& =\hfill & \frac{49}{2}\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill c& =\hfill & \pm \sqrt{\frac{49}{2}}\hfill \\ \text{Simplify the radical.}\hfill & & & \hfill c& =\hfill & \pm \frac{\sqrt{49}}{\sqrt{2}}\hfill \\ \text{Rationalize the denominator.}\hfill & & & \hfill c& =\hfill & \pm \frac{\sqrt{49}·\sqrt{2}}{\sqrt{2}·\sqrt{2}}\hfill \\ \text{Simplify.}\hfill & & & \hfill c& =\hfill & \pm \frac{7\sqrt{2}}{2}\hfill \end{array}\hfill \\ \\ \text{Rewrite to show two solutions.}c=\frac{7\sqrt{2}}{2},c=-\frac{7\sqrt{2}}{2}\hfill & & \\ \text{Check. We leave the check for you.}\hfill & & \end{array}$

Solution

	${(x-3)}^2=16$
Use the Square Root Property.	$x-3=\pm \sqrt{16}$
Simplify.	$x-3=\pm 4$
Write as two equations.	$x-3=4,x-3=\text{−}4$
Solve.	$x=7,x=\text{−}1$
Check.

Solution

	${(y-7)}^2=12$
Use the Square Root Property.	$y-7=\pm \sqrt{12}$
Simplify the radical.	$y-7=\pm 2\sqrt{3}$
Solve for y .	$y=7\pm 2\sqrt{3}$
Rewrite to show two solutions.	$y=7+2\sqrt{3},y=7-2\sqrt{3}$
Check.

Solution

$\begin{array}{c}\begin{array}{cccccc}& & & {\left(x-\frac{1}{2}\right)}^2\hfill & =\hfill & \frac{5}{4}\hfill \\ \text{Use the Square Root Property.}\hfill & & & x-\frac{1}{2}\hfill & =\hfill & \pm \sqrt{\frac{5}{4}}\hfill \\ \text{Rewrite the radical as a fraction of square roots.}\hfill & & & x-\frac{1}{2}\hfill & =\hfill & \pm \frac{\sqrt{5}}{\sqrt{4}}\hfill \\ \text{Simplify the radical.}\hfill & & & x-\frac{1}{2}\hfill & =\hfill & \pm \frac{\sqrt{5}}{2}\hfill \\ \text{Solve for}x.\hfill & & & \hfill x& =\hfill & \frac{1}{2}\pm \frac{\sqrt{5}}{2}\hfill \end{array}\hfill \\ \\ \text{Rewrite to show two solutions.}x=\frac{1}{2}+\frac{\sqrt{5}}{2},x=\frac{1}{2}-\frac{\sqrt{5}}{2}\hfill & & \\ \text{Check. We leave the check for you.}\hfill & & \end{array}$

Solution

$\begin{array}{c}\begin{array}{cccccc}& & & \hfill {\left(x-2\right)}^2+3& =\hfill & 30\hfill \\ \text{Isolate the binomial term.}\hfill & & & \hfill {\left(x-2\right)}^2& =\hfill & 27\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill x-2& =\hfill & \pm \sqrt{27}\hfill \\ \text{Simplify the radical.}\hfill & & & \hfill x-2& =\hfill & \pm 3\sqrt{3}\hfill \\ \text{Solve for}x.\hfill & & & \hfill x& =\hfill & 2\pm 3\sqrt{3}\hfill \end{array}\hfill \\ \\ \text{Rewrite to show two solutions.}x=2+3\sqrt{3},x=2-3\sqrt{3}\hfill \\ \text{Check. We leave the check for you.}\hfill & & \end{array}$

Solution

$\begin{array}{cccc}\text{Use the Square Root Property.}\hfill & & & \hfill \begin{array}{}\\ \\ \\ \hfill {\left(3v-7\right)}^2& =\hfill & \mathrm{-12}\hfill \\ \hfill 3v-7& =\hfill & \pm \sqrt{\mathrm{-12}}\hfill \end{array}\\ \text{The}\sqrt{\mathrm{-12}}\text{is not a real number.}\hfill & & & \hfill \text{There is no real solution.}\hfill \end{array}$