# Answer Key 10.4

1. $2^2-4(4)(-5)\Rightarrow 4+80=84\hspace{0.25in} \therefore 2\text{ real solutions}$
2. $(-6)^2-4(9)(1)\Rightarrow 36-36=0\hspace{0.25in} \therefore 1\text{ real solution}$
3. $(3)^2-4(2)(-5)\Rightarrow 9+40=49\hspace{0.25in} \therefore 2\text{ real solutions}$
4. $3x^2+5x-3\Rightarrow (5)^2-4(3)(-3)\Rightarrow 25+36=61\hspace{0.25in} \therefore 2\text{ real solutions}$
5. $3x^2+5x-2\Rightarrow (5)^2-4(3)(-2)\Rightarrow 25+24=49\hspace{0.25in} \therefore 2\text{ real solutions}$
6. $(-8)^2-4(1)(16)\Rightarrow 64-64=0\hspace{0.25in} \therefore 1\text{ real solution}$
7. $a^2+10a-56\Rightarrow (10)^2-4(1)(-56)\Rightarrow 100+224=324\hspace{0.25in} \therefore 2\text{ real solutions}$
8. $x^2-4x+4\Rightarrow (-4)^2-4(1)(4)\Rightarrow 16-16=0\hspace{0.25in} \therefore 1\text{ real solution}$
9. $5x^2-10x+26\Rightarrow (-10)^2-4(5)(26)\Rightarrow 100-520=-420$
$\therefore2\text{ non-real solutions}$
10. $n^2-10n+21\Rightarrow (-10)^2-4(1)(21)\Rightarrow 100-84=16$
$\therefore2\text{ real solutions}$

1. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}4 \\ b&=&\phantom{-}3 \\ c&=&-6 \\ \\ a&=&\dfrac{-3\pm \sqrt{3^2-4(4)(-6)}}{2(4)} \\ \\ a&=&\dfrac{-3\pm \sqrt{9+96}}{8} \\ \\ a&=&\dfrac{-3\pm \sqrt{105}}{8} \end{array}$
2. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}3 \\ b&=&\phantom{-}2 \\ c&=&-3 \\ \\ k&=&\dfrac{-2\pm \sqrt{2^2-4(3)(-3)}}{2(3)} \\ \\ k&=&\dfrac{-2\pm \sqrt{4+36}}{6} \\ \\ k&=&\dfrac{-2\pm \sqrt{40}}{6} \\ \\ k&=&\dfrac{-2\pm 2\sqrt{10}}{6} \Rightarrow \dfrac{-1\pm \sqrt{10}}{3} \end{array}$
3. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}2 \\ b&=&-8 \\ c&=&-2 \\ \\ x&=&\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(-2)}}{2(2)} \\ \\ x&=&\dfrac{8\pm \sqrt{64+16}}{4} \\ \\ x&=&\dfrac{8\pm \sqrt{80}}{4} \\ \\ x&=&\dfrac{8\pm 4\sqrt{5}}{4}\Rightarrow 2\pm \sqrt{5} \end{array}$
4. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}6 \\ b&=&\phantom{-}8 \\ c&=&-1 \\ \\ n&=&\dfrac{-8\pm \sqrt{8^2-4(6)(-1)}}{2(6)} \\ \\ n&=&\dfrac{-8\pm \sqrt{64+24}}{12} \\ \\ n&=&\dfrac{-8\pm \sqrt{88}}{12} \\ \\ n&=&\dfrac{-8\pm 2\sqrt{22}}{12}\Rightarrow \dfrac{-4\pm \sqrt{22}}{6} \end{array}$
5. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}2 \\ b&=&-3 \\ c&=&\phantom{-}6 \\ \\ m&=&\dfrac{-(-3)\pm \sqrt{(-3)^2-4(2)(6)}}{2(2)} \\ \\ m&=&\dfrac{3\pm \sqrt{9-48}}{4} \\ \\ m&=&\dfrac{3\pm \sqrt{-39}}{4} \\ \\ \end{array}$
A negative square root means there are 2 non-real solutions or no real solution.
6. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&5 \\ b&=&2 \\ c&=&6 \\ \\ p&=&\dfrac{-2\pm \sqrt{2^2-4(5)(6)}}{2(5)} \\ \\ p&=&\dfrac{-2\pm \sqrt{4-120}}{10} \\ \\ p&=&\dfrac{-2\pm \sqrt{-116}}{10} \end{array}$
A negative square root means there are 2 non-real solutions or no real solution.
7. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-}3 \\ b&=&-2 \\ c&=&-1 \\ \\ r&=&\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-1)}}{2(3)} \\ \\ r&=&\dfrac{2\pm \sqrt{4+12}}{6} \\ \\ r&=&\dfrac{2\pm \sqrt{16}}{6} \\ \\ r&=&\dfrac{2\pm 4}{6} \Rightarrow 1, -\dfrac{1}{3} \end{array}$
8. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{-0}2 \\ b&=&-\phantom{0}2 \\ c&=&-15 \\ \\ x&=&\dfrac{-(-2)\pm \sqrt{(-2)^2-4(2)(-15)}}{2(2)} \\ \\ x&=&\dfrac{2\pm \sqrt{4+120}}{4} \\ \\ x&=&\dfrac{2\pm \sqrt{124}}{4} \\ \\ x&=&\dfrac{2\pm 2\sqrt{31}}{4} \Rightarrow \dfrac{1\pm \sqrt{31}}{2} \end{array}$
9. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&\phantom{0}4 \\ b&=&-3 \\ c&=&10 \\ \\ n&=&\dfrac{-(-3)\pm \sqrt{(-3)^2-4(4)(10)}}{2(4)} \\ \\ n&=&\dfrac{3\pm \sqrt{9-160}}{8} \\ \\ n&=&\dfrac{3\pm \sqrt{-151}}{8} \\ \\ \end{array}$
$\therefore$ 2 non-real solutions
10. $\phantom{a}$
$\begin{array}[t]{rrl} a&=&1 \\ b&=&6 \\ c&=&9 \\ \\ b&=&\dfrac{-6\pm \sqrt{6^2-4(1)(9)}}{2(1)} \\ \\ b&=&\dfrac{-6\pm 0\cancel{\sqrt{36-36}}}{2} \\ \\ b&=&\dfrac{-6}{2}\Rightarrow -3 \end{array}$
11. $\phantom{a}$
$\begin{array}[t]{rrrrrrrrrrr} v^2&-&4v&-&5&=&-8&&&& \\ &&&+&8&&+8&&&& \\ \hline &&&&0&=&v^2&-&4v&+&3 \end{array}$$\begin{array}{rrl} a&=&\phantom{-}1 \\ b&=&-4 \\ c&=&\phantom{-}3 \\ \\ v&=&\dfrac{-(-4)\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)} \\ \\ v&=&\dfrac{4\pm \sqrt{16-12}}{2} \\ \\ v&=&\dfrac{4\pm \sqrt{4}}{2} \\ \\ v&=&\dfrac{4\pm 2}{2}\Rightarrow 2 \pm 1 \\ \\ v&=&3, 1 \end{array}$
12. $\phantom{a}$
$\begin{array}[t]{rrrrrrrrrrr} x^2&+&2x&+&6&=&4&&&& \\ &&&-&4&&-4&&&& \\ \hline &&&&0&=&x^2&+&2x&+&2 \end{array}\\ \begin{array}{rrl} a&=&1 \\ b&=&2 \\ c&=&2 \\ \\ x&=&\dfrac{-2\pm \sqrt{2^2-4(1)(2)}}{2(1)} \\ \\ x&=&\dfrac{-2\pm \sqrt{4-8}}{2} \\ \\ x&=&\dfrac{-2\pm \sqrt{-4}}{2} \end{array}$$\therefore$ 2 non-real solutions

## License

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.