2.1 Solving quadratic equations using the method of completing the  (Page 2/2)

$y^2+3y.$   The constant must be the square of one half the coefficient of $y.$ Since the coefficient of $y$ is 3, we have

$\begin{array}{lllllll}\frac{3}{2}\hfill & \hfill & \text{and}\hfill & \hfill & {\left(\frac{3}{2}\right)}^2\hfill & =\hfill & \frac{9}{4}\hfill \end{array}$
The constant is $\frac{9}{4}.$
$y^2+3y+\frac{9}{4}={\left(y+\frac{3}{2}\right)}^2$

$$\begin{array}{lllll}\hfill x^2+8x-9& =\hfill & 0.\hfill & \hfill & \text{Add\hspace{0.17em}9\hspace{0.17em}to\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\hfill \\ \hfill x^2+8x& =\hfill & 9\hfill & \hfill & \begin{array}{l}\text{One\hspace{0.17em}half\hspace{0.17em}the\hspace{0.17em}coefficient\hspace{0.17em}of\hspace{0.17em}}x{\text{\hspace{0.17em}is\hspace{0.17em}4,\hspace{0.17em}and\hspace{0.17em}4}}^{\text{2}}\text{\hspace{0.17em}is\hspace{0.17em}16}\text{.\hspace{0.17em}}\\ \text{Add\hspace{0.17em}16\hspace{0.17em}to\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\end{array}\hfill \\ \hfill x^2+8x+16& =\hfill & 9+16\hfill & \hfill & \hfill \\ \hfill x^2+8x+16& =\hfill & 25\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \hfill {\left(x+4\right)}^2& =\hfill & 25\hfill & \hfill & \text{Take\hspace{0.17em}square\hspace{0.17em}roots}\text{.}\hfill \\ \hfill x+4& =\hfill & \pm 5\hfill & \hfill & \hfill \\ \hfill x& =\hfill & \pm 5-4\hfill & \hfill & +5-4=1,-5-4=-9\hfill \\ \hfill x& =\hfill & 1,-9\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lllll}\hfill x^2-3x-1& =\hfill & 0.\hfill & \hfill & \text{Add\hspace{0.17em}1\hspace{0.17em}to\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\hfill \\ \hfill x^2-3x& =\hfill & 1\hfill & \hfill & \begin{array}{l}\text{One\hspace{0.17em}half\hspace{0.17em}the\hspace{0.17em}coefficient\hspace{0.17em}of\hspace{0.17em}}x\text{\hspace{0.17em}is\hspace{0.17em}}\frac{-3}{2}\text{.\hspace{0.17em}}\\ \text{Square\hspace{0.17em}it:\hspace{0.17em}}{\left(\frac{-3}{2}\right)}^{\text{2}}\text{=}\frac{9}{4}\text{.\hspace{0.17em}Add\hspace{0.17em}}\frac{9}{4}\text{\hspace{0.17em}to\hspace{0.17em}each\hspace{0.17em}side}\text{.}\end{array}\hfill \\ \hfill x^2-3x+\frac{9}{4}& =\hfill & 1+\frac{9}{4}\hfill & \hfill & \hfill \\ \hfill x^2-3x+\frac{9}{4}& =\hfill & \frac{13}{4}\hfill & \hfill & \begin{array}{l}\text{Factor}\text{.\hspace{0.17em}Notice\hspace{0.17em}that\hspace{0.17em}since\hspace{0.17em}the\hspace{0.17em}sign\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}middle\hspace{0.17em}}\\ \text{term\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}trinomial\hspace{0.17em}is\hspace{0.17em}"}-"\text{,\hspace{0.17em}its\hspace{0.17em}factored\hspace{0.17em}form\hspace{0.17em}has\hspace{0.17em}a\hspace{0.17em}"}-"\text{\hspace{0.17em}sign}\text{.}\end{array}\hfill \\ \hfill {\left(x-\frac{3}{2}\right)}^2& =\hfill & \frac{13}{4}\hfill & \hfill & \text{Now\hspace{0.17em}take\hspace{0.17em}square\hspace{0.17em}roots}\text{.}\hfill \\ \hfill x-\frac{3}{2}& =\hfill & \pm \sqrt{\frac{13}{4}}\hfill & \hfill & \hfill \\ \hfill x-\frac{3}{2}& =\hfill & \pm \sqrt{\frac{13}{2}}\hfill & \hfill & \hfill \\ \hfill x& =\hfill & \pm \frac{\sqrt{13}}{2}+\frac{3}{2}\hfill & \hfill & \hfill \\ \hfill x& =\hfill & \frac{\pm \sqrt{13}+3}{2}\hfill & \hfill & \hfill \\ \hfill x& =\hfill & \frac{3\pm \sqrt{13}}{2}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lllll}3a^2-36a-39\hfill & =\hfill & 0.\hfill & \hfill & \text{Add\hspace{0.17em}39\hspace{0.17em}to\hspace{0.17em}both\hspace{0.17em}sides}\text{.}\hfill \\ \hfill 3a^2-36a& =\hfill & 39\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}leading\hspace{0.17em}coefficient\hspace{0.17em}is\hspace{0.17em}3\hspace{0.17em}and\hspace{0.17em}we\hspace{0.17em}}\\ \text{need\hspace{0.17em}it\hspace{0.17em}to\hspace{0.17em}be\hspace{0.17em}1}\text{.\hspace{0.17em}Divide\hspace{0.17em}each\hspace{0.17em}term\hspace{0.17em}by\hspace{0.17em}3}\text{.}\end{array}\hfill \\ \hfill a^2-12a& =\hfill & 13\hfill & \hfill & \begin{array}{l}\text{One\hspace{0.17em}half\hspace{0.17em}the\hspace{0.17em}coefficient\hspace{0.17em}of\hspace{0.17em}}a\text{\hspace{0.17em}is\hspace{0.17em}}-6.\\ \text{Square\hspace{0.17em}it:\hspace{0.17em}}{\left(-6\right)}^{\text{2}}\text{=36}\text{.\hspace{0.17em}Add\hspace{0.17em}36\hspace{0.17em}to\hspace{0.17em}each\hspace{0.17em}side}\text{.}\end{array}\hfill \\ \hfill a^2-12a+36& =\hfill & 13+36\hfill & \hfill & \hfill \\ \hfill a^2-12a+36& =\hfill & 49\hfill & \hfill & \hfill \\ \hfill {\left(a-6\right)}^2& =\hfill & 49\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \hfill a-6& =\hfill & \pm 7\hfill & \hfill & \hfill \\ \hfill a& =\hfill & \pm 7+6\hfill & \hfill & +7+6=13,-7+6=-1\hfill \\ \hfill a=13,& \hfill & -1\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lll}\hfill 2x^2+x+4& =\hfill & 0\hfill \\ \hfill 2x^2+x& =\hfill & -4\hfill \\ \hfill x^2+\frac{1}{2}x& =\hfill & -2\hfill \\ \hfill x^2+\frac{1}{2}x+{\left(\frac{1}{4}\right)}^2& =\hfill & -2+{\left(\frac{1}{4}\right)}^2\hfill \\ {\left(x+\frac{1}{4}\right)}^2=-2+\frac{1}{16}\hfill & =\hfill & \frac{-32}{16}+\frac{1}{16}=\frac{-31}{16}\hfill \end{array}$$
Since we know that the square of any number is positive, this equation has no real number solution.

 Calculator problem.  Solve $7a^2-5a-1=0.$ Round each solution to the nearest tenth.
$$\begin{array}{lllllll}\hfill 7a^2-5a-1& =\hfill & 0\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill 7a^2-5a& =\hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill a^2-\frac{5}{7}a& =\hfill & \frac{1}{7}\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill a^2-\frac{5}{7}a+{\left(\frac{5}{14}\right)}^2& =\hfill & \frac{1}{7}+{\left(\frac{5}{14}\right)}^2\hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill {\left(a-\frac{5}{14}\right)}^2& =\hfill & \frac{1}{7}+\frac{25}{196}\hfill & =\hfill & \frac{28}{196}+\frac{25}{196}\hfill & =\hfill & \frac{53}{196}\hfill \\ \hfill a-\frac{5}{14}& =\hfill & \pm \sqrt{\frac{53}{196}}\hfill & =\hfill & \pm \frac{\sqrt{53}}{14}\hfill & \hfill & \hfill \\ \hfill a=\frac{5}{14}\pm \frac{\sqrt{53}}{14}& =\hfill & \frac{5\pm \sqrt{53}}{14}\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$

1. We will first compute the value of the square root.
   
   $\begin{array}{lll}\text{Type}\hfill & \hfill & 53\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline \sqrt{x}\\ \hline\end{array}\hfill \\ \text{Display\hspace{0.17em}reads:}\hfill & \hfill & 7.2801099\hfill \end{array}$
   Press the key that places this value into memory.
2. For $a=\frac{5+\sqrt{53}}{14},$
   
   $\begin{array}{l}\begin{array}{lll}\text{Type}\hfill & \hfill & 5\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline +\\ \hline\end{array}\hfill \end{array}\\ \text{Press\hspace{0.17em}the\hspace{0.17em}key\hspace{0.17em}that\hspace{0.17em}recalls\hspace{0.17em}the\hspace{0.17em}value\hspace{0.17em}in\hspace{0.17em}memory}\text{.}\\ \begin{array}{lll}\text{Press}\hfill & \hfill & \begin{array}{||}\hline =\\ \hline\end{array}\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline ÷\\ \hline\end{array}\hfill \\ \text{Type}\hfill & \hfill & 14\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline =\\ \hline\end{array}\hfill \\ \text{Display\hspace{0.17em}reads:}\hfill & \hfill & .87715071\hfill \end{array}\end{array}$
   Rounding to tenths, we get $a\approx \mathrm{0.9.}$
3. For $a=\frac{5-\sqrt{53}}{14}$
   
   $\begin{array}{l}\begin{array}{lll}\text{Type}\hfill & \hfill & 5\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline -\\ \hline\end{array}\hfill \end{array}\\ \text{Press\hspace{0.17em}the\hspace{0.17em}key\hspace{0.17em}that\hspace{0.17em}recalls\hspace{0.17em}the\hspace{0.17em}value\hspace{0.17em}in\hspace{0.17em}memory}\text{.}\\ \begin{array}{lll}\text{Press}\hfill & \hfill & \begin{array}{||}\hline =\\ \hline\end{array}\hfill \\ \text{Press}\hfill & \hfill & \begin{array}{||}\hline ÷\\ \hline\end{array}\hfill \\ \text{Type}\hfill & \hfill & 14\hfill \\ \text{Display\hspace{0.17em}reads:}\hfill & \hfill & -.16286499\hfill \end{array}\end{array}$
   Rounding to tenths, we get $a\approx -\mathrm{0.2.}$ Thus, $a\approx 0.9$ and $-0.2$ to the nearest tenth.