Back to the class
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 $x=\dfrac{1}{2}$.