Test 12 (Lessons 23–24): Solving Quadratic Equations Solutions

1. ${\left(2x+7\right)}^2=20$

$\begin{array}{rcl}\sqrt{{\left(2x+7\right)}^2}& =& \pm \sqrt{20}\\ 2x+7& =& \pm 2\sqrt{5}\\ 2x& =& –7\pm 2\sqrt{5}\end{array}$

$x=\frac{–7\pm 2\sqrt{5}}{2}$

2. $6x^2+2x=0$

$$\begin{gathered} 2x\left(3x+1\right)=0 \\ \begin{array}{cc}2x=0,& 3x+1=0\end{array} \end{gathered}$$

$x=–\frac{1}{3}, 0$

3. Given the solutions, without determining the exact equation, explain the degree you would expect the equation to have.
4. $x=\pm 15i, –\frac{3}{8}$

$\begin{array}{ccc}x=–15i,& x=15i,& x=–\frac{3}{8}\end{array}$

Sample: The equation has three solutions. The number of solutions determines the degree of the polynomial equation. Therefore, the equation would be a third degree polynomial.

4. $x^2 +\frac{5}{3}x + c$

$$\begin{gathered} \begin{array}{ccc}c={\left(\frac{b}{2}\right)}^2& a=1,& b=\frac{5}{3}\end{array} \\ c={\left(\frac{5}{2·3}\right)}^2 \end{gathered}$$

$c=\frac{25}{36}$

5. $x^2 + bx + \frac{121}{4}$

$$\begin{gathered} \begin{array}{ccc}b=2\sqrt{c}& a=1,& c=\frac{121}{4}\end{array} \\ b=2\sqrt{\frac{121}{4}}=2\left(\frac{11}{2}\right) \end{gathered}$$

$b=11$

6. Solve by completing the square.
   $2x^2+6x+7=0$

$\begin{array}{rcl}\frac{2}{2}{x}^2+\frac{6}{2}x+\frac{7}{2}& =& 0\\ x^2+3x& =& –\frac{7}{2}\\ x^2+3x+\boxed{ {\left(\frac{3}{2}\right)}^2 }& =& –\frac{7}{2}+\boxed{ {\left(\frac{3}{2}\right)}^2 }\\ {\left(x+\frac{3}{2}\right)}^2& =& –\frac{7}{2}+\frac{9}{4}\\ {\left(x+\frac{3}{2}\right)}^2& =& –\frac{14}{4}+\frac{9}{4}\\ {\left(x+\frac{3}{2}\right)}^2& =& –\frac{5}{4}\\ \sqrt{{\left(x+\frac{3}{2}\right)}^2}& =& \pm \sqrt{–\frac{5}{4}}\\ x+\frac{3}{2}& =& \frac{\pm i\sqrt{5}}{2}\end{array}$

$x=\frac{–3\pm i\sqrt{5}}{2} \mathrm{or} x=–\frac{3}{2}\pm \frac{i\sqrt{5}}{2}$

7. Name any missing roots. Then determine a possible polynomial equation to represent the roots.
   $i, –\sqrt{7}$

$\begin{array}{rcl}& & \begin{array}{cccccc}i,& –i,& \sqrt{7},& –\sqrt{7}& & \\ x=i,& x=–i,& x=\sqrt{7},& x=–\sqrt{7}& & \end{array}\end{array}$

$\begin{array}{rcl}\left(x–i\right)\left(x–\left(–i\right)\right)\left(x–\sqrt{7}\right)\left(x–\left(–\sqrt{7}\right)\right)& =& 0\\ \left(x–i\right)\left(x+i\right)\left(x–\sqrt{7}\right)\left(x+\sqrt{7}\right)& =& 0\\ \left(x^2–i^2\right)\left(x^2–\sqrt{49}\right)& =& 0\\ \left(x^2–\left(–1\right)\right)\left(x^2–7\right)& =& 0\\ \left(x^2+1\right)\left(x^2–7\right)& =& 0\\ x^4–7x^2+x^2–7& =& 0\end{array}$

The missing roots are: $–i, \sqrt{7},$ a possible polynomial equation: $x^4–6x^2–7=0$

8. Solve by completing the square.
   $x^2–8x=–4$

$\begin{array}{rcl}{x}^2–8x+\boxed{{\left(–\frac{8}{2}\right)}^2 }& =& –4+\boxed{{\left(–\frac{8}{2}\right)}^2 }\\ {\left(x–4\right)}^2& =& –4+16\\ {\left(x–4\right)}^2& =& 12\\ \sqrt{{\left(x–4\right)}^2}& =& \pm \sqrt{12}\\ x–4& =& \pm 2\sqrt{3}\end{array}$

$\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & \pm \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \sqrt{3}\end{array}$