College Algebra Quiz: Quadratic Formula And The Discriminant
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Quadratic Formula And The DiscriminantQuestion 1 of 5

The equation kx2+8x+k=0kx^2 + 8x + k = 0 has real solutions when the discriminant is non-negative. For which values of kk does this equation have real solutions?

k≤−4k \leq -4 or k≥4k \geq 4
−4≤k≤4-4 \leq k \leq 4
k≤−4k \leq -4 or k>0k > 0
k<0k < 0 or k≥4k \geq 4
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College Algebra Quiz

College Algebra Quiz: Quadratic Formula And The Discriminant

Practice Quadratic Formula And The Discriminant 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 Quadratic Formula And The Discriminant, 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.

All questions

Question 1

The equation kx2+8x+k=0kx^2 + 8x + k = 0 has real solutions when the discriminant is non-negative. For which values of kk does this equation have real solutions?

  1. k≤−4k \leq -4 or k≥4k \geq 4
  2. −4≤k≤4-4 \leq k \leq 4 (correct answer)
  3. k≤−4k \leq -4 or k>0k > 0
  4. k<0k < 0 or k≥4k \geq 4
Explanation: For real solutions, we need Δ≥0\Delta \geq 0. Here a=ka = k, b=8b = 8, c=kc = k, so Δ=82−4(k)(k)=64−4k2\Delta = 8^2 - 4(k)(k) = 64 - 4k^2. Setting 64−4k2≥064 - 4k^2 \geq 0: 64≥4k264 \geq 4k^2, so 16≥k216 \geq k^2, which means −4≤k≤4-4 \leq k \leq 4. We also need k≠0k \neq 0 for this to be quadratic, but the question asks about real solutions to the equation as given. Choice A gives the complement of the correct interval. Choice C incorrectly excludes some valid kk values. Choice D has an incorrect boundary.

Question 2

Consider the family of quadratics fk(x)=x2−6x+kf_k(x) = x^2 - 6x + k where kk is a parameter. For which value of kk will the minimum value of fk(x)f_k(x) be equal to −4-4?

  1. k=−4k = -4
  2. k=9k = 9
  3. k=13k = 13
  4. k=5k = 5 (correct answer)
Explanation: When you encounter a quadratic function and need to find its minimum value, remember that parabolas opening upward (positive leading coefficient) have their minimum at the vertex. The key is finding the vertex and using it to determine the parameter. For fk(x)=x2−6x+kf_k(x) = x^2 - 6x + k, you can find the vertex using the formula x=−b2ax = -\frac{b}{2a}. Here, a=1a = 1 and b=−6b = -6, so the x-coordinate of the vertex is x=−(−6)2(1)=3x = -\frac{(-6)}{2(1)} = 3. This means the minimum occurs at x=3x = 3 regardless of the value of kk. The minimum value is fk(3)=32−6(3)+k=9−18+k=k−9f_k(3) = 3^2 - 6(3) + k = 9 - 18 + k = k - 9. Since we want this minimum to equal −4-4, we set up the equation: k−9=−4k - 9 = -4, which gives us k=5k = 5. Looking at the wrong answers: Choice A (k=−4k = -4) gives a minimum of −4−9=−13-4 - 9 = -13, not −4-4. This might tempt you if you mistakenly think the minimum value equals kk directly. Choice B (k=9k = 9) produces a minimum of 9−9=09 - 9 = 0, which occurs when the parabola just touches the x-axis. Choice C (k=13k = 13) yields a minimum of 13−9=413 - 9 = 4, the positive version of our target. Study tip: For vertex form problems, always find where the minimum occurs first (the x-coordinate), then substitute back to find the actual minimum value. The parameter kk shifts the parabola vertically, but the vertex's x-coordinate stays the same.

Question 3

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c has discriminant Δ=25\Delta = 25. If the function has roots at x=2x = 2 and x=7x = 7, what is the value of aa?

  1. a=1a = 1 (correct answer)
  2. a=15a = \frac{1}{5}
  3. a=5a = 5
  4. a=25a = 25
Explanation: If the roots are x=2x = 2 and x=7x = 7, then f(x)=a(x−2)(x−7)=a(x2−9x+14)=ax2−9ax+14af(x) = a(x-2)(x-7) = a(x^2 - 9x + 14) = ax^2 - 9ax + 14a. So b=−9ab = -9a and c=14ac = 14a. The discriminant is Δ=b2−4ac=(−9a)2−4a(14a)=81a2−56a2=25a2\Delta = b^2 - 4ac = (-9a)^2 - 4a(14a) = 81a^2 - 56a^2 = 25a^2. Given Δ=25\Delta = 25, we have 25a2=2525a^2 = 25, so a2=1a^2 = 1, giving a=±1a = \pm 1. Since we need a specific value and typically take the positive case unless specified otherwise, a=1a = 1. Choice B would result from confusing the discriminant formula. Choice C comes from 25=5\sqrt{25} = 5. Choice D uses the discriminant value directly.

Question 4

For what values of kk will the equation 2x2−8x+k=02x^2 - 8x + k = 0 have exactly one real solution?

  1. k=8k = 8 (correct answer)
  2. k=4k = 4
  3. k=16k = 16
  4. k=−8k = -8
Explanation: For a quadratic equation to have exactly one real solution, the discriminant must equal zero. Using ax2+bx+c=0ax^2 + bx + c = 0, we have a=2a = 2, b=−8b = -8, and c=kc = k. The discriminant is b2−4ac=(−8)2−4(2)(k)=64−8kb^2 - 4ac = (-8)^2 - 4(2)(k) = 64 - 8k. Setting this equal to zero: 64−8k=064 - 8k = 0, so k=8k = 8. Choice B uses c=2kc = 2k instead of kk. Choice C incorrectly uses b2b^2 as the discriminant. Choice D uses the wrong sign.

Question 5

The quadratic 2x2+px+8=02x^2 + px + 8 = 0 has discriminant Δ=p2−64\Delta = p^2 - 64. For what value of pp will the equation have a repeated root, and what will that root be?

  1. p=8p = 8; root is x=−2x = -2
  2. p=−8p = -8; root is x=2x = 2
  3. p=±8p = \pm 8; root is x=∓2x = \mp 2 (correct answer)
  4. p=16p = 16; root is x=−4x = -4
Explanation: For a repeated root, the discriminant must equal zero: p2−64=0p^2 - 64 = 0, so p2=64p^2 = 64, giving p=±8p = \pm 8. When p=8p = 8: the equation is 2x2+8x+8=02x^2 + 8x + 8 = 0 or x2+4x+4=0x^2 + 4x + 4 = 0, which factors as (x+2)2=0(x+2)^2 = 0, so x=−2x = -2. When p=−8p = -8: the equation is 2x2−8x+8=02x^2 - 8x + 8 = 0 or x2−4x+4=0x^2 - 4x + 4 = 0, which factors as (x−2)2=0(x-2)^2 = 0, so x=2x = 2. Thus p=±8p = \pm 8 gives roots x=∓2x = \mp 2 respectively. Choice A only considers one case. Choice B only considers the other case. Choice D uses an incorrect value of pp.