Answer Key 10.4
[latexpage]
- \(2^2-4(4)(-5)\Rightarrow 4+80=84\hspace{0.25in} \therefore 2\text{ real solutions}\)
- \((-6)^2-4(9)(1)\Rightarrow 36-36=0\hspace{0.25in} \therefore 1\text{ real solution}\)
- \((3)^2-4(2)(-5)\Rightarrow 9+40=49\hspace{0.25in} \therefore 2\text{ real solutions}\)
- \(3x^2+5x-3\Rightarrow (5)^2-4(3)(-3)\Rightarrow 25+36=61\hspace{0.25in} \therefore 2\text{ real solutions}\)
- \(3x^2+5x-2\Rightarrow (5)^2-4(3)(-2)\Rightarrow 25+24=49\hspace{0.25in} \therefore 2\text{ real solutions}\)
- \((-8)^2-4(1)(16)\Rightarrow 64-64=0\hspace{0.25in} \therefore 1\text{ real solution}\)
- \(a^2+10a-56\Rightarrow (10)^2-4(1)(-56)\Rightarrow 100+224=324\hspace{0.25in} \therefore 2\text{ real solutions}\)
- \(x^2-4x+4\Rightarrow (-4)^2-4(1)(4)\Rightarrow 16-16=0\hspace{0.25in} \therefore 1\text{ real solution}\)
- \(5x^2-10x+26\Rightarrow (-10)^2-4(5)(26)\Rightarrow 100-520=-420\)
∴ \(2\text{ non-real solutions}\) - \(n^2-10n+21\Rightarrow (-10)^2-4(1)(21)\Rightarrow 100-84=16\)
∴ \(2\text{ real solutions}\)
- \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}4 \\
b&=&\phantom{-}3 \\
c&=&-6 \\ \\
a&=&\dfrac{-3\pm \sqrt{3^2-4(4)(-6)}}{2(4)} \\ \\
a&=&\dfrac{-3\pm \sqrt{9+96}}{8} \\ \\
a&=&\dfrac{-3\pm \sqrt{105}}{8}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}3 \\
b&=&\phantom{-}2 \\
c&=&-3 \\ \\
k&=&\dfrac{-2\pm \sqrt{2^2-4(3)(-3)}}{2(3)} \\ \\
k&=&\dfrac{-2\pm \sqrt{4+36}}{6} \\ \\
k&=&\dfrac{-2\pm \sqrt{40}}{6} \\ \\
k&=&\dfrac{-2\pm 2\sqrt{10}}{6} \Rightarrow \dfrac{-1\pm \sqrt{10}}{3}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}2 \\
b&=&-8 \\
c&=&-2 \\ \\
x&=&\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(-2)}}{2(2)} \\ \\
x&=&\dfrac{8\pm \sqrt{64+16}}{4} \\ \\
x&=&\dfrac{8\pm \sqrt{80}}{4} \\ \\
x&=&\dfrac{8\pm 4\sqrt{5}}{4}\Rightarrow 2\pm \sqrt{5}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}6 \\
b&=&\phantom{-}8 \\
c&=&-1 \\ \\
n&=&\dfrac{-8\pm \sqrt{8^2-4(6)(-1)}}{2(6)} \\ \\
n&=&\dfrac{-8\pm \sqrt{64+24}}{12} \\ \\
n&=&\dfrac{-8\pm \sqrt{88}}{12} \\ \\
n&=&\dfrac{-8\pm 2\sqrt{22}}{12}\Rightarrow \dfrac{-4\pm \sqrt{22}}{6}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}2 \\
b&=&-3 \\
c&=&\phantom{-}6 \\ \\
m&=&\dfrac{-(-3)\pm \sqrt{(-3)^2-4(2)(6)}}{2(2)} \\ \\
m&=&\dfrac{3\pm \sqrt{9-48}}{4} \\ \\
m&=&\dfrac{3\pm \sqrt{-39}}{4} \\ \\
\end{array}\)A negative square root means there are 2 non-real solutions or no real solution.
- \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&5 \\
b&=&2 \\
c&=&6 \\ \\
p&=&\dfrac{-2\pm \sqrt{2^2-4(5)(6)}}{2(5)} \\ \\
p&=&\dfrac{-2\pm \sqrt{4-120}}{10} \\ \\
p&=&\dfrac{-2\pm \sqrt{-116}}{10}
\end{array}\)A negative square root means there are 2 non-real solutions or no real solution.
- \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-}3 \\
b&=&-2 \\
c&=&-1 \\ \\
r&=&\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-1)}}{2(3)} \\ \\
r&=&\dfrac{2\pm \sqrt{4+12}}{6} \\ \\
r&=&\dfrac{2\pm \sqrt{16}}{6} \\ \\
r&=&\dfrac{2\pm 4}{6} \Rightarrow 1, -\dfrac{1}{3}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{-0}2 \\
b&=&-\phantom{0}2 \\
c&=&-15 \\ \\
x&=&\dfrac{-(-2)\pm \sqrt{(-2)^2-4(2)(-15)}}{2(2)} \\ \\
x&=&\dfrac{2\pm \sqrt{4+120}}{4} \\ \\
x&=&\dfrac{2\pm \sqrt{124}}{4} \\ \\
x&=&\dfrac{2\pm 2\sqrt{31}}{4} \Rightarrow \dfrac{1\pm \sqrt{31}}{2}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&\phantom{0}4 \\
b&=&-3 \\
c&=&10 \\ \\
n&=&\dfrac{-(-3)\pm \sqrt{(-3)^2-4(4)(10)}}{2(4)} \\ \\
n&=&\dfrac{3\pm \sqrt{9-160}}{8} \\ \\
n&=&\dfrac{3\pm \sqrt{-151}}{8} \\ \\
\end{array}\)∴ 2 non-real solutions
- \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
a&=&1 \\
b&=&6 \\
c&=&9 \\ \\
b&=&\dfrac{-6\pm \sqrt{6^2-4(1)(9)}}{2(1)} \\ \\
b&=&\dfrac{-6\pm 0\cancel{\sqrt{36-36}}}{2} \\ \\
b&=&\dfrac{-6}{2}\Rightarrow -3
\end{array}\) - \(\begin{array}{rrrrrrrrrrr}
\\ \\
v^2&-&4v&-&5&=&-8&&&& \\
&&&+&8&&+8&&&& \\
\midrule
&&&&0&=&v^2&-&4v&+&3
\end{array}\)\(\begin{array}{rrl}
a&=&\phantom{-}1 \\
b&=&-4 \\
c&=&\phantom{-}3 \\ \\
v&=&\dfrac{-(-4)\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)} \\ \\
v&=&\dfrac{4\pm \sqrt{16-12}}{2} \\ \\
v&=&\dfrac{4\pm \sqrt{4}}{2} \\ \\
v&=&\dfrac{4\pm 2}{2}\Rightarrow 2 \pm 1 \\ \\
v&=&3, 1
\end{array}\) - \(\begin{array}{rrrrrrrrrrr}
\\ \\
x^2&+&2x&+&6&=&4&&&& \\
&&&-&4&&-4&&&& \\
\midrule
&&&&0&=&x^2&+&2x&+&2
\end{array}\)\(\begin{array}{rrl}
a&=&1 \\
b&=&2 \\
c&=&2 \\ \\
x&=&\dfrac{-2\pm \sqrt{2^2-4(1)(2)}}{2(1)} \\ \\
x&=&\dfrac{-2\pm \sqrt{4-8}}{2} \\ \\
x&=&\dfrac{-2\pm \sqrt{-4}}{2}
\end{array}\)∴ 2 non-real solutions