College Algebra Quiz: Complex Solutions To Quadratic Equations

What this quiz covers

This quiz focuses on Complex Solutions To Quadratic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 3
2. 5
3. 7 (correct answer)
4. 1

Explanation: Since the coefficients are real, complex solutions come in conjugate pairs. If $-1 + 2i$ is a solution, then $-1 - 2i$ is also a solution. The quadratic is $(x - (-1 + 2i))(x - (-1 - 2i)) = (x + 1 - 2i)(x + 1 + 2i) = ((x + 1) - 2i)((x + 1) + 2i) = (x + 1)^2 - (2i)^2 = (x + 1)^2 + 4 = x^2 + 2x + 1 + 4 = x^2 + 2x + 5$. So $b = 2$ and $c = 5$, giving $b + c = 7$. Choice A forgets to square the imaginary part correctly. Choice B uses $c$ alone. Choice D uses $b - c$ instead.

Question 2

1. 4
2. 9
3. 13 (correct answer)
4. -5

Explanation: If the solutions are $2 + 3i$ and $2 - 3i$, then the quadratic can be written as $a(x - (2 + 3i))(x - (2 - 3i)) = 0$. Expanding: $a((x - 2) - 3i)((x - 2) + 3i) = a((x - 2)^2 - (3i)^2) = a((x - 2)^2 + 9) = a(x^2 - 4x + 4 + 9) = a(x^2 - 4x + 13)$. Since $a = 1$, we have $x^2 - 4x + 13 = 0$, so $c = 13$. Choice A uses only the real part squared. Choice B uses only the imaginary part squared. Choice D incorrectly subtracts instead of adding the squares.

Question 3

1. 8
2. 9 (correct answer)
3. 16
4. 17

Explanation: For complex solutions, the discriminant $b^2 - 4ac < 0$. Here $a = 2$, $b = -8$, $c = k$. So $(-8)^2 - 4(2)(k) < 0$, which gives $64 - 8k < 0$, so $k > 8$. The smallest integer greater than 8 is 9. Choice A gives $k = 8$, making the discriminant zero (one repeated real solution). Choice C uses $k = 16$ but ignores that we need the smallest integer. Choice D incorrectly calculates the discriminant condition.

Question 4

1. 18
2. 19 (correct answer)
3. 17
4. 20

Explanation: Using the quadratic formula: $x = \frac{12 \pm \sqrt{144 - 8k}}{4} = \frac{12 \pm \sqrt{144 - 8k}}{4} = 3 \pm \frac{\sqrt{144 - 8k}}{4}$. For the real part to be 3, we need the solutions to be $3 \pm bi$ for some real $b \neq 0$. This means $\frac{\sqrt{144 - 8k}}{4}$ must be purely imaginary, so $144 - 8k < 0$, giving $k > 18$. The smallest integer greater than 18 is 19. Choice A gives $k = 18$, which makes the discriminant zero (one repeated real solution). Choice C gives $k = 17 < 18$. Choice D is larger than necessary.

Question 5

1. 25 (correct answer)
2. -7
3. 9
4. 16

Explanation: Since $p$ and $q$ are real, the solutions must be complex conjugates. So $\beta = 3 + 4i$. The product of the solutions is $\alpha \cdot \beta = (3 - 4i)(3 + 4i) = 9 - (4i)^2 = 9 - 16i^2 = 9 + 16 = 25$. Alternatively, by Vieta's formulas, $\alpha \cdot \beta = q$, and we can find $q$ from the factored form: $(x - (3 - 4i))(x - (3 + 4i)) = (x - 3 + 4i)(x - 3 - 4i) = ((x - 3) + 4i)((x - 3) - 4i) = (x - 3)^2 + 16 = x^2 - 6x + 9 + 16 = x^2 - 6x + 25$. So $q = 25$. Choice B uses the sum instead of product. Choice C uses only the real part squared. Choice D uses only the imaginary part squared.

Question 6

1. 2 (correct answer)
2. $2\sqrt{2}$
3. 4
4. $2i$

Explanation: Using the quadratic formula: $r_1, r_2 = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2} = -3 \pm i$. So $r_1 = -3 + i$ and $r_2 = -3 - i$. Then $r_1 - r_2 = (-3 + i) - (-3 - i) = 2i$, and $|r_1 - r_2| = |2i| = 2$. Choice B incorrectly uses the distance formula in the complex plane. Choice C doubles the correct answer. Choice D gives the complex number itself rather than its modulus.