College Algebra Quiz: Complex Solutions To Quadratic Equations
6 questions · exam conditions
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Complex Solutions To Quadratic EquationsQuestion 1 of 6

If z=−1+2iz = -1 + 2i is one solution to the quadratic equation x2+bx+c=0x^2 + bx + c = 0 where bb and cc are real numbers, what is the value of b+cb + c?

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College Algebra Quiz

College Algebra Quiz: Complex Solutions To Quadratic Equations

Practice Complex Solutions To Quadratic Equations in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

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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.

All questions

Question 1

If z=−1+2iz = -1 + 2i is one solution to the quadratic equation x2+bx+c=0x^2 + bx + c = 0 where bb and cc are real numbers, what is the value of b+cb + c?

  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-1 + 2i is a solution, then −1−2i-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=x2+2x+1+4=x2+2x+5(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=2b = 2 and c=5c = 5, giving b+c=7b + c = 7. Choice A forgets to square the imaginary part correctly. Choice B uses cc alone. Choice D uses b−cb - c instead.

Question 2

A quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0 has solutions x=2±3ix = 2 \pm 3i. If a=1a = 1, what is the value of cc?

  1. 4
  2. 9
  3. 13 (correct answer)
  4. -5
Explanation: If the solutions are 2+3i2 + 3i and 2−3i2 - 3i, then the quadratic can be written as a(x−(2+3i))(x−(2−3i))=0a(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(x2−4x+4+9)=a(x2−4x+13)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=1a = 1, we have x2−4x+13=0x^2 - 4x + 13 = 0, so c=13c = 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

The quadratic equation 2x2−8x+k=02x^2 - 8x + k = 0 has complex solutions when the discriminant is negative. If the equation has exactly one complex conjugate pair of solutions, what is the smallest integer value of kk that satisfies this condition?

  1. 8
  2. 9 (correct answer)
  3. 16
  4. 17
Explanation: For complex solutions, the discriminant b2−4ac<0b^2 - 4ac < 0. Here a=2a = 2, b=−8b = -8, c=kc = k. So (−8)2−4(2)(k)<0(-8)^2 - 4(2)(k) < 0, which gives 64−8k<064 - 8k < 0, so k>8k > 8. The smallest integer greater than 8 is 9. Choice A gives k=8k = 8, making the discriminant zero (one repeated real solution). Choice C uses k=16k = 16 but ignores that we need the smallest integer. Choice D incorrectly calculates the discriminant condition.

Question 4

The equation 2x2−12x+k=02x^2 - 12x + k = 0 has two complex solutions. If the real parts of both solutions are equal to 3, what is the smallest possible integer value of kk such that the solutions are non-real?

  1. 18
  2. 19 (correct answer)
  3. 17
  4. 20
Explanation: Using the quadratic formula: x=12±144−8k4=12±144−8k4=3±144−8k4x = \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±bi3 \pm bi for some real b≠0b \neq 0. This means 144−8k4\frac{\sqrt{144 - 8k}}{4} must be purely imaginary, so 144−8k<0144 - 8k < 0, giving k>18k > 18. The smallest integer greater than 18 is 19. Choice A gives k=18k = 18, which makes the discriminant zero (one repeated real solution). Choice C gives k=17<18k = 17 < 18. Choice D is larger than necessary.

Question 5

The polynomial P(x)=x2+6x+10P(x) = x^2 + 6x + 10 can be written in the form P(x)=(x−r1)(x−r2)P(x) = (x - r_1)(x - r_2) where r1r_1 and r2r_2 are complex numbers. What is ∣r1−r2∣|r_1 - r_2|?

  1. 2 (correct answer)
  2. 222\sqrt{2}
  3. 4
  4. 2i2i
Explanation: Using the quadratic formula: r1,r2=−6±36−402=−6±−42=−6±2i2=−3±ir_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 r1=−3+ir_1 = -3 + i and r2=−3−ir_2 = -3 - i. Then r1−r2=(−3+i)−(−3−i)=2ir_1 - r_2 = (-3 + i) - (-3 - i) = 2i, and ∣r1−r2∣=∣2i∣=2|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.

Question 6

The quadratic x2+px+q=0x^2 + px + q = 0 has solutions α\alpha and β\beta. If α=3−4i\alpha = 3 - 4i, and pp and qq are real, what is the value of α⋅β\alpha \cdot \beta?

  1. 25 (correct answer)
  2. -7
  3. 9
  4. 16
Explanation: Since pp and qq are real, the solutions must be complex conjugates. So β=3+4i\beta = 3 + 4i. The product of the solutions is α⋅β=(3−4i)(3+4i)=9−(4i)2=9−16i2=9+16=25\alpha \cdot \beta = (3 - 4i)(3 + 4i) = 9 - (4i)^2 = 9 - 16i^2 = 9 + 16 = 25. Alternatively, by Vieta's formulas, α⋅β=q\alpha \cdot \beta = q, and we can find qq 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=x2−6x+9+16=x2−6x+25(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=25q = 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.