10.3 Solving quadratic equations using the method of extraction of

$\begin{array}{cccccc}{x}^2-49\hfill & =\hfill & 0.\hfill & \text{Rewrite}\text{.}\hfill & \hfill & \hfill \\ \hfill x^2& =\hfill & 49\hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & \pm \sqrt{49}\hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & \pm 7\hfill & \hfill & \hfill & \hfill \\ Check:\hfill & \hfill & {(7)}^2=49\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & {(-7)}^2=49\hfill & \text{Is this correct}\hfill \\ \hfill & \hfill & 49=49\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill & 49=49\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$

$\begin{array}{rrr}25a^2& =& 36\hfill \\ a^2& =& \frac{36}{25}\hfill \\ a& =& \pm \sqrt{\frac{36}{25}}\hfill \\ a& =& \pm \frac{6}{5}\hfill \end{array}$
$\begin{array}{lllllllllll}Check:\hfill & \hfill & \hfill 25{(\frac{6}{5})}^2& =\hfill & 36\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill & \hfill 25{(\frac{-6}{5})}^2& =\hfill & 36\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 25{(\frac{36}{25})}^2& =\hfill & 36\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill & \hfill 25(\frac{36}{25})& =\hfill & 36\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 36& =\hfill & 36\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill & \hfill & \hfill 36& =\hfill & 36\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$

$\begin{array}{rrr}4m^2-32& =& 0\hfill \\ 4m^2& =\hfill & 32\hfill \\ m^2& =& \frac{32}{4}\hfill \\ m^2& =& 8\hfill \\ m& =& \pm \sqrt{8}\hfill \\ m& =& \pm 2\sqrt{2}\hfill \end{array}$
$\begin{array}{llllllllll}\hfill Check:& \hfill & \hfill 4{(2\sqrt{2})}^2& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 4{(-2\sqrt{2})}^2& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 4[2^2{(\sqrt{2})}^2]& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 4[{(-2)}^2{(\sqrt{2})}^2]& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 4[4·2]& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 4[4·2]& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 4·8& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 4·8& =\hfill & 32\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 32& =\hfill & 32\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill & \hfill 32& =\hfill & 32\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$

Solve $5x^2-15y^2{z}^7=0$ for $x.$
$\begin{array}{lllll}5x^2& =& 15y^2{z}^7& & \text{Divide\hspace{0.17em}both\hspace{0.17em}sides\hspace{0.17em}by\hspace{0.17em}5}\text{.}\\ \hfill x^2& =& 3y^2{z}^7& & \\ \hfill x& =& \pm \sqrt{3y^2{z}^7}& & \\ \hfill x& =& \pm y{z}^3\sqrt{3z}& & \end{array}$

Calculator problem.  Solve $14a^2-235=0.$ Round to the nearest hundredth.
$\begin{array}{ccccc}14a^2-235\hfill & =& 0.\hfill & & \text{Rewrite}\text{.}\hfill \\ \hfill 14a^2& =& 235& & \text{Divide\hspace{0.17em}both\hspace{0.17em}sides\hspace{0.17em}by\hspace{0.17em}14}\text{.}\\ \hfill a^2& =& \frac{235}{14}& & \end{array}$
$\text{On\hspace{0.17em}the\hspace{0.17em}Calculator}$
$\begin{array}{ccc}\text{Type}& & 235\\ \text{Press}& & \begin{array}{|c|}\hline ÷\\ \hline\end{array}\\ \text{Type}& & 14\\ \text{Press}& & \begin{array}{|c|}\hline =\\ \hline\end{array}\\ \text{Press}& & \surd \\ \text{Display\hspace{0.17em}reads:}& & 4.0970373\end{array}$
Rounding to the nearest hundredth produces 4.10. We must be sure to insert the $\pm$ symbol. $a\approx \pm 4.10$

$\begin{array}{lll}{k}^2& =& -64\\ k& =& \pm \sqrt{-64}\end{array}$
The radicand is negative so no real number solutions exist.

$$\begin{array}{lllllllll}\hfill {(x+2)}^2& =\hfill & 81\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x+2& =\hfill & \pm \sqrt{81}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x+2& =\hfill & \pm 9\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Subtract\hspace{0.17em}}2\text{\hspace{0.17em}from\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\hfill \\ \hfill x& =\hfill & -2\pm 9\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & -2+9\hfill & \hfill & \text{and}\hfill & \hfill & \hfill x& =\hfill & -2-9\hfill \\ \hfill x& =\hfill & 7\hfill & \hfill & \hfill & \hfill & \hfill x& =\hfill & -11\hfill \end{array}$$

$$\begin{array}{lllll}{\left(a+3\right)}^2& =& 5& & \\ \hfill a+3& =& \pm \sqrt{5}& & \text{Subtract\hspace{0.17em}3\hspace{0.17em}from\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\\ \hfill a& =& -3\pm \sqrt{5}& & \end{array}$$