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Section 3.1 #2: Express $\sqrt{-21}$ in terms of $i$.
Solution: Compute
$$\sqrt{-21} = \sqrt{21}i.$$
Section 3.1 #12: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers:$$(-6-5i)+(9+2i).$$
Solution: Compute
$$(-6-5i)+(9+2i)=(-6+9)+(-5i+2i)=3-3i.$$
Section 3.1 #39: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers:
$$(1+3i)(1-4i).$$
Solution: Compute
$$(1+3i)(1-4i)=1-4i+3i-12i^2=1-i+12=13-i$$
Section 3.1 #74: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers:
$$\dfrac{\sqrt{5}+3i}{1-i}.$$
Solution: Compute
$$\begin{array}{ll}
\dfrac{\sqrt{5}+3i}{1-i} &= \left( \dfrac{\sqrt{5}+3i}{1-i} \right) \left( \dfrac{1+i}{1+i} \right) \\
&= \dfrac{\sqrt{5}+\sqrt{5}i+3i+3i^2}{1-i^2} \\
&= \dfrac{\sqrt{5} + (\sqrt{5}+3)i-3}{1-(-1)} \\
&= \dfrac{(\sqrt{5}-3)+(\sqrt{5}+3)i}{2} \\
&= \dfrac{\sqrt{5}-3}{2} + \dfrac{\sqrt{5}+3}{2} i.
\end{array}$$
Section 3.1 #85: Simplify
$$(-i)^{71}.$$
Solution: We want to exploit the fact that $i^2=-1$. First rewrite this as
$$(-i)^{71}=(-1)^{71} i^{71} = (-1)i i^{70}.$$
To finish, we compute
$$(-i)^{71} = -i i^{70} = -i \left( i^2 \right)^{35} = -i (-1)^{35} = -i (-1) = i.$$
Section 3.2 #34: Solve the quadratic equation
$$x^2+6x+13=0.$$
Solution: Use the quadratic formula with $a=1$, $b=6$, and $c=13$ to compute
$$\begin{array}{ll}
x&=\dfrac{-6 \pm \sqrt{6^2-4(1)(13)}}{2}\\
&=\dfrac{-6 \pm \sqrt{36-52}}{2}\\
&=\dfrac{-6 \pm \sqrt{-16}}{2} \\
&= \dfrac{-6 \pm 4i}{2} \\
&= -3 \pm 2i.
\end{array}$$
Section 3.2 #42: Solve the quadratic equation
$$3t^2+8t+3=0.$$
Solution: Use the quadratic formula iwht $a=3$, $b=8$, and $c=3$ to compute
$$\begin{array}{ll}
x &= \dfrac{-8 \pm \sqrt{8^2 - 4(3)(3)}}{2(3)} \\
&= \dfrac{-8 \pm \sqrt{64 - 36}}{6} \\
&= \dfrac{-8 \pm \sqrt{28}}{6} \\
&= \dfrac{-8 \pm 2\sqrt{7}}{6} \\
&= -\dfrac{4}{3} \pm \dfrac{\sqrt{7}}{3}.
\end{array}$$
Section 3.3 #4: Find the vertex, find the axis of symmetry, determine whether there is a maximum or a minimum value (and find it), and graph the function
$$g(x)=x^2+7x-8.$$
Solution: The coefficient of $x^2$ is already $1$ so to complete the square, add and subtract $\left( \dfrac{7}{2} \right)^2$ and then factor to get
$$\begin{array}{ll}
x^2+7x-8&=x^2+7x-8 + \left( \dfrac{7}{2} \right)^2 - \left( \dfrac{7}{2} \right)^2 \\
&=\left( x+ \dfrac{7}{2} \right)^2 - 8 - \left( \dfrac{7}{2} \right)^2 \\
&=\left( x + \dfrac{7}{2} \right)^2 - \dfrac{32}{4} - \dfrac{49}{4} \\
&=\left( x+ \dfrac{7}{2} \right)^2 - \dfrac{81}{4}.
\end{array}$$
From this we see that the vertex is $\left(-\dfrac{7}{2}, -\dfrac{81}{4} \right)$. The axis of symmetry is $x=-\dfrac{7}{2}$. There is no maximum. The minimum value is $-\dfrac{81}{4}$ and it occurs as $x=-\dfrac{7}{2}$.
Section 3.3 #15: Find the vertex, find the axis of symmetry, determine whether there is a maximum or a minimum value (and find it), and graph the function
$$g(x)=-2x^2+2x+1.$$
Solution: First factor out the $-2$ to get
$$-2\left(x^2-x-\dfrac{1}{2} \right).$$
Inside, we will complete the square by adding and subtracting $\left( - \dfrac{1}{2} \right)^2$:
$$\begin{array}{ll}
-2\left( x^2-x - \dfrac{1}{2} \right) &= -2 \left( x^2 - x - \dfrac{1}{2} + \left( - \dfrac{1}{2} \right)^2 - \left( -\dfrac{1}{2} \right)^2 \right) \\
&= -2 \left( \left( x-\dfrac{1}{2} \right)^2 - \dfrac{1}{2} - \dfrac{1}{4} \right) \\
&= -2 \left( \left( x -\dfrac{1}{2} \right)^2 - \dfrac{3}{4} \right) \\
&= -2 \left( x-\dfrac{1}{2} \right)^2 + \dfrac{6}{4} \\
&= -2 \left( x - \dfrac{1}{2} \right)^2 + \dfrac{3}{2}.
\end{array}$$
From this we see the vertex is $\left( \dfrac{1}{2}, \dfrac{3}{2} \right)$. The axis of symmetry is $x= \dfrac{1}{2}$, there is no minimum, and the maximum is $\dfrac{3}{2}$ which occurs at $x=\dfrac{1}{2}$.
