ACCUPLACER Advanced Algebra & Functions Quiz: Discriminant And Solution Types
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Discriminant And Solution TypesQuestion 1 of 20

How many distinct real solutions does the equation 13x2+x+23=0\frac{1}{3}x^2 + x + \frac{2}{3} = 0 have?

None
One
Two
Three
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Discriminant And Solution Types

Practice Discriminant And Solution Types in ACCUPLACER Advanced Algebra & Functions 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 Discriminant And Solution Types, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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

How many distinct real solutions does the equation 13x2+x+23=0\frac{1}{3}x^2 + x + \frac{2}{3} = 0 have?

  1. None
  2. One
  3. Two (correct answer)
  4. Three
Explanation: To simplify the calculation, we can clear the fractions by multiplying the entire equation by the least common denominator, 3. This gives x2+3x+2=0x^2 + 3x + 2 = 0. Now we find the discriminant, D=b2−4acD = b^2 - 4ac, with a=1a=1, b=3b=3, and c=2c=2. D=(3)2−4(1)(2)=9−8=1D = (3)^2 - 4(1)(2) = 9 - 8 = 1. Since the discriminant is positive (D>0D > 0), the equation has two distinct real solutions.

Question 2

The discriminant of a quadratic equation with integer coefficients is 20. Which of the following correctly describes the solutions?

  1. One repeated rational solution
  2. Two distinct rational solutions
  3. Two distinct irrational solutions (correct answer)
  4. Two non-real solutions
Explanation: The nature of the solutions is determined by the discriminant, DD. Since D=20D = 20 is positive, the equation has two distinct real solutions. To determine if they are rational or irrational, we check if the discriminant is a perfect square. Since 20 is not a perfect square, the solutions will be irrational.

Question 3

Which of the following quadratic equations has no real solutions?

  1. x2−8x+16=0x^2 - 8x + 16 = 0
  2. 2x2+5x−3=02x^2 + 5x - 3 = 0
  3. 3x2−x+1=03x^2 - x + 1 = 0 (correct answer)
  4. x2−7=0x^2 - 7 = 0
Explanation: An equation has no real solutions if its discriminant D=b2−4acD = b^2 - 4ac is negative. We check each option: A) D=(−8)2−4(1)(16)=64−64=0D = (-8)^2 - 4(1)(16) = 64 - 64 = 0. B) D=(5)2−4(2)(−3)=25+24=49>0D = (5)^2 - 4(2)(-3) = 25 + 24 = 49 > 0. C) D=(−1)2−4(3)(1)=1−12=−11<0D = (-1)^2 - 4(3)(1) = 1 - 12 = -11 < 0. D) D=(0)2−4(1)(−7)=28>0D = (0)^2 - 4(1)(-7) = 28 > 0. Only option C has a negative discriminant, so it is the correct answer.

Question 4

Describe the nature of the roots of the equation −x2+8x−16=0-x^2 + 8x - 16 = 0.

  1. Two distinct irrational roots
  2. Two distinct rational roots
  3. One repeated rational root (correct answer)
  4. Two non-real roots
Explanation: To find the nature of the roots, we compute the discriminant D=b2−4acD = b^2 - 4ac. For −x2+8x−16=0-x^2 + 8x - 16 = 0, we have a=−1a=-1, b=8b=8, and c=−16c=-16. The discriminant is D=(8)2−4(−1)(−16)=64−64=0D = (8)^2 - 4(-1)(-16) = 64 - 64 = 0. A discriminant of 0 indicates that there is exactly one real solution, which is a repeated rational root.

Question 5

For what values of pp does the equation x2+(p−1)x+4=0x^2 + (p-1)x + 4 = 0 have exactly one real solution?

  1. p=3p = 3 and p=−5p = -5
  2. p=5p = 5 and p=−3p = -3 (correct answer)
  3. p=4p = 4 and p=−4p = -4
  4. p=5p = 5 only
Explanation: For exactly one real solution, the discriminant must be zero. Here, a=1a=1, b=p−1b = p-1, and c=4c=4. Setting D=b2−4ac=0D = b^2 - 4ac = 0 gives (p−1)2−4(1)(4)=0(p-1)^2 - 4(1)(4) = 0. This simplifies to (p−1)2−16=0(p-1)^2 - 16 = 0, so (p−1)2=16(p-1)^2 = 16. Taking the square root of both sides gives p−1=±4p-1 = \pm 4. This leads to two possible values for pp: p=1+4=5p = 1 + 4 = 5 and p=1−4=−3p = 1 - 4 = -3.

Question 6

Find the set of all non-zero values of kk for which the equation kx2+6x+1=0kx^2 + 6x + 1 = 0 has two distinct real solutions.

  1. k>9k > 9
  2. k<9k < 9 and k≠0k \neq 0 (correct answer)
  3. k<−9k < -9
  4. k>−9k > -9 and k≠0k \neq 0
Explanation: For two distinct real solutions, the discriminant D=b2−4acD = b^2 - 4ac must be greater than 0. Here, a=ka=k, b=6b=6, and c=1c=1. So, we have D=(6)2−4(k)(1)>0D = (6)^2 - 4(k)(1) > 0. This simplifies to 36−4k>036 - 4k > 0. Adding (4k) to both sides gives 36>4k36 > 4k. Dividing by 4 gives 9>k9 > k, or k<9k < 9. The question specifies that kk is non-zero, so the full condition is k<9k < 9 and k≠0k \neq 0.

Question 7

If a quadratic equation ax2+bx+c=0ax^2+bx+c=0 with integer coefficients has a discriminant of 1, which of the following is true about its solutions?

  1. There is one repeated irrational solution.
  2. There are two distinct irrational solutions.
  3. There are no real solutions.
  4. There are two distinct rational solutions. (correct answer)
Explanation: The discriminant D=1D=1. Since D>0D>0, there are two distinct real solutions. Since D=1D=1 is a perfect square (12=11^2=1), the solutions are rational. Therefore, the equation has two distinct rational solutions.

Question 8

What is the nature of the solutions for the equation 3x(x−2)=−43x(x - 2) = -4?

  1. One repeated real solution
  2. Two distinct rational solutions
  3. Two distinct irrational solutions
  4. Two non-real solutions (correct answer)
Explanation: First, put the equation into the standard form ax2+bx+c=0ax^2 + bx + c = 0. Distributing on the left gives 3x2−6x=−43x^2 - 6x = -4. Adding 4 to both sides gives 3x2−6x+4=03x^2 - 6x + 4 = 0. Now, calculate the discriminant, D=b2−4acD = b^2 - 4ac, with a=3a=3, b=−6b=-6, and c=4c=4. D=(−6)2−4(3)(4)=36−48=−12D = (-6)^2 - 4(3)(4) = 36 - 48 = -12. Since the discriminant is negative, the equation has two non-real (complex) solutions.

Question 9

Which of the following best describes the solutions to the equation 4x2−9=04x^2 - 9 = 0?

  1. Two distinct rational solutions (correct answer)
  2. Two distinct irrational solutions
  3. One repeated rational solution
  4. No real solutions
Explanation: In the equation 4x2−9=04x^2 - 9 = 0, we have a=4a=4, b=0b=0, and c=−9c=-9. The discriminant is D=b2−4ac=(0)2−4(4)(−9)=0+144=144D = b^2 - 4ac = (0)^2 - 4(4)(-9) = 0 + 144 = 144. Since the discriminant is positive (D>0D > 0), there are two distinct real solutions. Because the discriminant (144) is a perfect square (12212^2), the solutions are rational.

Question 10

The graph of a quadratic function y=ax2+bx+cy = ax^2 + bx + c is a parabola that is tangent to the x-axis. What is the value of the discriminant b2−4acb^2 - 4ac?

  1. It is a negative value.
  2. It is a positive value.
  3. It is equal to 1.
  4. It is equal to 0. (correct answer)
Explanation: When the graph of a quadratic function is tangent to the x-axis, it means the parabola touches the x-axis at exactly one point (the vertex). This corresponds to the quadratic equation having exactly one real solution (a repeated root). This occurs precisely when the discriminant, b2−4acb^2 - 4ac, is equal to 0.

Question 11

Which of the following values for the discriminant of a quadratic equation with rational coefficients corresponds to an equation with two distinct, irrational roots?

  1. -1
  2. 0
  3. 24 (correct answer)
  4. 49
Explanation: For a quadratic equation to have two distinct, irrational roots, its discriminant must be positive (for two real roots) and not be a perfect square (for irrational roots). A) -1 is negative (no real roots). B) 0 results in one repeated rational root. C) 24 is positive and not a perfect square, so it corresponds to two distinct irrational roots. D) 49 is a perfect square (727^2), so it corresponds to two distinct rational roots.

Question 12

The equation x2−6x+p=0x^2 - 6x + p = 0 has discriminant Δ1\Delta_1, and the equation x2+6x+p=0x^2 + 6x + p = 0 has discriminant Δ2\Delta_2. Which statement is always true?

  1. Δ1>Δ2\Delta_1 > \Delta_2 for all real values of pp
  2. Δ1<Δ2\Delta_1 < \Delta_2 for all real values of pp
  3. Δ1=Δ2\Delta_1 = \Delta_2 for all real values of pp (correct answer)
  4. The relationship between Δ1\Delta_1 and Δ2\Delta_2 depends on the value of pp
Explanation: For the first equation: Δ1=(−6)2−4(1)(p)=36−4p\Delta_1 = (-6)^2 - 4(1)(p) = 36 - 4p. For the second equation: Δ2=(6)2−4(1)(p)=36−4p\Delta_2 = (6)^2 - 4(1)(p) = 36 - 4p. Therefore Δ1=Δ2\Delta_1 = \Delta_2 for all real values of pp. The discriminant depends on b2b^2, and (−6)2=62(-6)^2 = 6^2. Choices A and B incorrectly suggest one is always larger. Choice D incorrectly suggests the relationship varies with pp.

Question 13

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has discriminant Δ=25\Delta = 25. If the equation is modified by replacing cc with c+6c + 6, what can be concluded about the number of real solutions of the new equation?

  1. The new equation definitely has two distinct real solutions
  2. The new equation definitely has no real solutions
  3. The new equation definitely has exactly one real solution
  4. The number of real solutions cannot be determined without more information (correct answer)
Explanation: The original discriminant is Δ1=b2−4ac=25\Delta_1 = b^2 - 4ac = 25. The new discriminant is Δ2=b2−4a(c+6)=b2−4ac−24a=25−24a\Delta_2 = b^2 - 4a(c + 6) = b^2 - 4ac - 24a = 25 - 24a. Since we don't know the value of aa, we cannot determine the sign of Δ2\Delta_2. If a>2524a > \frac{25}{24}, then Δ2<0\Delta_2 < 0 (no real solutions). If a=2524a = \frac{25}{24}, then Δ2=0\Delta_2 = 0 (one real solution). If 0<a<25240 < a < \frac{25}{24}, then Δ2>0\Delta_2 > 0 (two real solutions).

Question 14

For what positive value of pp does the equation x2+px+4p=0x^2 + px + 4p = 0 have exactly one real solution?

  1. 4
  2. 8
  3. 16 (correct answer)
  4. 32
Explanation: For the equation to have exactly one real solution, its discriminant must be zero. Here, a=1a=1, b=pb=p, and c=4pc=4p. So we set D=b2−4ac=0D = b^2 - 4ac = 0. Substituting the coefficients gives p2−4(1)(4p)=0p^2 - 4(1)(4p) = 0, which simplifies to p2−16p=0p^2 - 16p = 0. Factoring out pp gives p(p−16)=0p(p - 16) = 0. The solutions are p=0p=0 and p=16p=16. The question asks for the positive value, which is 16.

Question 15

For what value of kk does the quadratic equation x2−10x+k=0x^2 - 10x + k = 0 have exactly one real solution?

  1. -25
  2. 5
  3. 25 (correct answer)
  4. 100
Explanation: A quadratic equation has exactly one real solution when its discriminant, D=b2−4acD = b^2 - 4ac, is equal to 0. For the equation x2−10x+k=0x^2 - 10x + k = 0, we have a=1a=1, b=−10b=-10, and c=kc=k. Setting the discriminant to zero: D=(−10)2−4(1)(k)=0D = (-10)^2 - 4(1)(k) = 0. This simplifies to 100−4k=0100 - 4k = 0. Solving for kk gives 4k=1004k = 100, so k=25k = 25.

Question 16

The equation 3x2+6x−k=03x^2 + 6x - k = 0 has two distinct real solutions. Which of the following inequalities must be true?

  1. k>−3k > -3 (correct answer)
  2. k<−3k < -3
  3. k>3k > 3
  4. k<3k < 3
Explanation: For an equation to have two distinct real solutions, the discriminant, D=b2−4acD = b^2 - 4ac, must be greater than 0. Here, a=3a=3, b=6b=6, and c=−kc=-k. So, we must have D=(6)2−4(3)(−k)>0D = (6)^2 - 4(3)(-k) > 0. This simplifies to 36+12k>036 + 12k > 0. Subtracting 36 from both sides gives 12k>−3612k > -36. Dividing by 12 gives k>−3k > -3.

Question 17

Find the set of all values of mm for which the equation x2−5x+m=0x^2 - 5x + m = 0 has no real solutions.

  1. m<254m < \frac{25}{4}
  2. m>254m > \frac{25}{4} (correct answer)
  3. m=254m = \frac{25}{4}
  4. m>−254m > -\frac{25}{4}
Explanation: An equation has no real solutions if its discriminant is negative (D<0D < 0). For the equation x2−5x+m=0x^2 - 5x + m = 0, we have a=1a=1, b=−5b=-5, and c=mc=m. The discriminant is D=(−5)2−4(1)(m)=25−4mD = (-5)^2 - 4(1)(m) = 25 - 4m. We set this to be less than 0: 25−4m<025 - 4m < 0. Adding (4m) to both sides gives 25<4m25 < 4m. Dividing by 4 gives 254<m\frac{25}{4} < m, or m>254m > \frac{25}{4}.

Question 18

How many distinct real solutions does the equation 2x2−7x+8=02x^2 - 7x + 8 = 0 have?

  1. None (correct answer)
  2. One
  3. Two
  4. Infinitely many
Explanation: To determine the number of distinct real solutions, we use the discriminant, D=b2−4acD = b^2 - 4ac, for the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. Here, a=2a=2, b=−7b=-7, and c=8c=8. The discriminant is D=(−7)2−4(2)(8)=49−64=−15D = (-7)^2 - 4(2)(8) = 49 - 64 = -15. Since the discriminant is negative (D<0D < 0), the equation has no real solutions; it has two complex conjugate solutions.

Question 19

The equation x2+bx+1=0x^2+bx+1=0 has no real solutions. Which of the following must be true for bb?

  1. b>2b > 2 or b<−2b < -2
  2. b=2b = 2 or b=−2b = -2
  3. b<−2b < -2
  4. −2<b<2-2 < b < 2 (correct answer)
Explanation: For no real solutions, the discriminant must be negative. Here a=1,c=1a=1, c=1. D=b2−4(1)(1)<0D = b^2 - 4(1)(1) < 0, which means b2−4<0b^2 - 4 < 0, or b2<4b^2 < 4. Taking the square root of both sides, we get b2<4\sqrt{b^2} < \sqrt{4}, which is ∣b∣<2|b| < 2. This inequality is equivalent to −2<b<2-2 < b < 2.

Question 20

A quadratic equation with rational coefficients has two distinct real solutions. Which condition guarantees that these solutions are also rational?

  1. The discriminant is any positive number.
  2. The discriminant is a perfect square greater than zero. (correct answer)
  3. The discriminant is an integer but not a perfect square.
  4. The discriminant is equal to zero.
Explanation: The solutions of a quadratic equation are given by the quadratic formula, which involves D\sqrt{D}, where D is the discriminant. For the solutions to be rational, D\sqrt{D} must be rational. This occurs when D is a perfect square. Since the solutions must also be distinct and real, D must be positive. Therefore, the discriminant must be a perfect square greater than zero.