Complex Numbers

$16 - 4(1)(13) = -36$.

• Correct: $-36$:
• $36$:
• $68$:
• $-52$:

A difference of squares: the middle terms cancel.

• Correct: $x^2 - 9$:
• $x^2 + 9$:
• $x^2 - 6x - 9$:
• $x^2 + 6x - 9$:

$i = \sqrt{-1}$, so $\sqrt{-16} = \sqrt{16}\,i = 4i$. Check: $(4i)^2 = 16i^2 = -16$.

• $-4$: $(-4)^2 = 16$, not $-16$. No real number squares to a negative.
• Correct: $4i$: Right: $\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i$, and $(4i)^2 = 16i^2 = -16$.
• $-4i$: $(-4i)^2$ is also $-16$, but $\sqrt{-16}$ names the root $4i$.
• $16i$: Take the square root of $16$ as well: $4i$.

Distribute: $3 \cdot 1$$+ 3(-4i)$$+ 2i \cdot 1$$+ 2i(-4i)$$= 3 - 12i$$+ 2i - 8i^2$. Since $i^2 = -1$, this is $3 + 8 - 10i = 11 - 10i$.

• Correct: $11 - 10i$: Right: $3 - 12i$$+ 2i - 8i^2$$= 3 - 10i$$+ 8$$= 11 - 10i$.
• $-5 - 10i$: $-8i^2 = -8(-1) = +8$, so the real part is $3 + 8 = 11$.
• $3 - 8i$: This multiplies the real parts, $3 \cdot 1 = 3$, and the imaginary parts' coefficients, $2 \cdot (-4) = -8$, separately. Distribute all four products, then use $i^2 = -1$.
• $11 + 14i$: The $i$ terms are $-12i + 2i = -10i$.

$b^2 - 4ac = 16 - 52 = -36$, and $\sqrt{-36} = 6i$. So $x$$= \frac{-4 \pm 6i}{2}$$= -2 \pm 3i$.

• $x = 2 \pm 3i$: The formula starts with $-b = -4$, so the real part is $-2$.
• $x = -2 \pm 3$: $\sqrt{-36} = 6i$, not $6$; the solutions are not real.
• $x = -4 \pm 6i$: Divide every term by $2a = 2$: $\frac{-4 \pm 6i}{2} = -2 \pm 3i$.
• Correct: $x = -2 \pm 3i$: Right: $\frac{-4 \pm \sqrt{16 - 52}}{2}$$= \frac{-4 \pm 6i}{2}$.

The complex conjugate of $a + bi$ is $a - bi$: same real part, opposite imaginary part.

• $-5 + 2i$: Only the sign of the imaginary part changes; the real part stays $5$.
• Correct: $5 + 2i$: Right: change the sign of the imaginary part.
• $-5 - 2i$: That changes the real part instead of the imaginary part.
• $2 - 5i$: The parts do not swap places.

$(a - bi)(a + bi)$$= a^2 + b^2$: $25 - 10i + 10i - 4i^2$$= 25 + 4$$= 29$. A number times its conjugate is real.

• Answer: 29

$b^2 - 4ac$$= 16 - 52$$= -36 < 0$: no real solutions, and the quadratic formula gives the pair $-2 \pm 3i$. For real coefficients, such solutions come as a number and its complex conjugate.

• Correct: None real; two complex solutions that are complex conjugates: Right: the discriminant is $-36$, and the solutions $-2 + 3i$ and $-2 - 3i$ are conjugates.
• Two real solutions, $-2$ and $3$: $-2$ and $3$ are the parts of $-2 + 3i$, not separate solutions.
• One real solution, $-2$, where the parabola turns: $x = -2$ is where the vertex is; the parabola never reaches the $x$-axis.
• Zero solutions of any kind, since the parabola stays above the axis: It has no real solutions, but it does have two complex ones.