ACCUPLACER Advanced Algebra & Functions Quiz: Quadratic Formula
20 questions · exam conditions
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Quadratic FormulaQuestion 1 of 20

A quadratic equation is given by 3x(x−2)=43x(x - 2) = 4. What are the solutions for xx?

3±213\frac{3 \pm \sqrt{21}}{3}
−3±213\frac{-3 \pm \sqrt{21}}{3}
1±1331 \pm \frac{\sqrt{13}}{3}
2,−232, -\frac{2}{3}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Quadratic Formula

Practice Quadratic Formula 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 Quadratic Formula, 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

A quadratic equation is given by 3x(x−2)=43x(x - 2) = 4. What are the solutions for xx?

  1. 3±213\frac{3 \pm \sqrt{21}}{3} (correct answer)
  2. −3±213\frac{-3 \pm \sqrt{21}}{3}
  3. 1±1331 \pm \frac{\sqrt{13}}{3}
  4. 2,−232, -\frac{2}{3}
Explanation: First, rewrite the equation in the standard form ax2+bx+c=0ax^2 + bx + c = 0. Distribute 3x3x to get 3x2−6x=43x^2 - 6x = 4. Then, subtract 4 from both sides: 3x2−6x−4=03x^2 - 6x - 4 = 0. Now, use the quadratic formula with a=3a=3, b=−6b=-6, and c=−4c=-4. x=−(−6)±(−6)2−4(3)(−4)2(3)=6±36+486=6±846x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-4)}}{2(3)} = \frac{6 \pm \sqrt{36 + 48}}{6} = \frac{6 \pm \sqrt{84}}{6}. Simplify the radical: 84=4⋅21=221\sqrt{84} = \sqrt{4 \cdot 21} = 2\sqrt{21}. So, x=6±2216x = \frac{6 \pm 2\sqrt{21}}{6}. Divide each term in the numerator and the denominator by 2: x=3±213x = \frac{3 \pm \sqrt{21}}{3}.

Question 2

The solutions to ax2+bx+c=0ax^2 + bx + c = 0 are x=5±174x = \frac{5 \pm \sqrt{17}}{4}. Which of the following could be the quadratic equation?

  1. 4x2−5x+17=04x^2 - 5x + 17 = 0
  2. 2x2+5x+1=02x^2 + 5x + 1 = 0
  3. 2x2−5x+1=02x^2 - 5x + 1 = 0 (correct answer)
  4. 4x2−10x+2=04x^2 - 10x + 2 = 0
Explanation: Comparing the given solution to the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}, we can deduce the coefficients. From the denominator, 2a=42a = 4, so a=2a = 2. From the numerator, −b=5-b = 5, so b=−5b = -5. The term under the radical is the discriminant, so b2−4ac=17b^2 - 4ac = 17. Substituting the known values of aa and bb: (−5)2−4(2)c=17⇒25−8c=17⇒8c=8⇒c=1(-5)^2 - 4(2)c = 17 \Rightarrow 25 - 8c = 17 \Rightarrow 8c = 8 \Rightarrow c = 1. Therefore, the equation is 2x2−5x+1=02x^2 - 5x + 1 = 0.

Question 3

For what positive value of kk will the equation kx2−12x+k=0kx^2 - 12x + k = 0 have exactly one real solution?

  1. 3
  2. 6 (correct answer)
  3. 12
  4. 36
Explanation: An equation has exactly one real solution when its discriminant, b2−4acb^2 - 4ac, is equal to 0. For the equation kx2−12x+k=0kx^2 - 12x + k = 0, we have a=ka = k, b=−12b = -12, and c=kc = k. Setting the discriminant to zero: (−12)2−4(k)(k)=0⇒144−4k2=0⇒4k2=144⇒k2=36(-12)^2 - 4(k)(k) = 0 \Rightarrow 144 - 4k^2 = 0 \Rightarrow 4k^2 = 144 \Rightarrow k^2 = 36. The possible values for kk are k=6k=6 and k=−6k=-6. The question asks for the positive value, which is 6.

Question 4

What is the product of the solutions to the equation 5x2−9x=35x^2 - 9x = 3?

  1. −95-\frac{9}{5}
  2. −35-\frac{3}{5} (correct answer)
  3. 35\frac{3}{5}
  4. 95\frac{9}{5}
Explanation: First, put the equation into standard form ax2+bx+c=0ax^2 + bx + c = 0: 5x2−9x−3=05x^2 - 9x - 3 = 0. For any quadratic equation in standard form, the product of the solutions (roots) is given by the formula ca\frac{c}{a}. In this equation, a=5a=5, b=−9b=-9, and c=−3c=-3. Therefore, the product of the solutions is −35\frac{-3}{5}. The sum of the solutions would be −ba=−−95=95-\frac{b}{a} = -\frac{-9}{5} = \frac{9}{5}.

Question 5

If x1x_1 and x2x_2 are the roots of the equation 3x2−9x+4=03x^2 - 9x + 4 = 0, what is the value of 1x1+1x2\frac{1}{x_1} + \frac{1}{x_2}?

  1. 49\frac{4}{9}
  2. 34\frac{3}{4}
  3. 33
  4. 94\frac{9}{4} (correct answer)
Explanation: To find the sum of the reciprocals 1x1+1x2\frac{1}{x_1} + \frac{1}{x_2}, we can use a common denominator: x2+x1x1x2\frac{x_2 + x_1}{x_1x_2}. This expression is the sum of the roots divided by the product of the roots. For the equation ax2+bx+c=0ax^2+bx+c=0, the sum of the roots is −ba-\frac{b}{a} and the product is ca\frac{c}{a}. For 3x2−9x+4=03x^2 - 9x + 4 = 0, the sum is −−93=3-\frac{-9}{3} = 3. The product is 43\frac{4}{3}. Therefore, 1x1+1x2=sumproduct=34/3=3⋅34=94\frac{1}{x_1} + \frac{1}{x_2} = \frac{\text{sum}}{\text{product}} = \frac{3}{4/3} = 3 \cdot \frac{3}{4} = \frac{9}{4}.

Question 6

What is the positive difference between the two solutions of the equation 2x2−6x+1=02x^2 - 6x + 1 = 0?

  1. 33
  2. 272\sqrt{7}
  3. 7\sqrt{7} (correct answer)
  4. 72\frac{\sqrt{7}}{2}
Explanation: First, find the two solutions using the quadratic formula: x=−(−6)±(−6)2−4(2)(1)2(2)=6±36−84=6±284x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(1)}}{2(2)} = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4}. Simplifying the radical, 28=27\sqrt{28} = 2\sqrt{7}. So, x=6±274=3±72x = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2}. The two solutions are x1=3+72x_1 = \frac{3 + \sqrt{7}}{2} and x2=3−72x_2 = \frac{3 - \sqrt{7}}{2}. Their difference is x1−x2=(3+72)−(3−72)=3+7−3+72=272=7x_1 - x_2 = \left(\frac{3 + \sqrt{7}}{2}\right) - \left(\frac{3 - \sqrt{7}}{2}\right) = \frac{3 + \sqrt{7} - 3 + \sqrt{7}}{2} = \frac{2\sqrt{7}}{2} = \sqrt{7}.

Question 7

For the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the value of the discriminant is DD. Which expression represents the sum of the two solutions?

  1. −b2a\frac{-b}{2a}
  2. −ba-\frac{b}{a} (correct answer)
  3. Da\frac{\sqrt{D}}{a}
  4. ca\frac{c}{a}
Explanation: The two solutions given by the quadratic formula are x1=−b+D2ax_1 = \frac{-b + \sqrt{D}}{2a} and x2=−b−D2ax_2 = \frac{-b - \sqrt{D}}{2a}, where D=b2−4acD = b^2 - 4ac. To find their sum, we add them: x1+x2=(−b+D2a)+(−b−D2a)=−b+D−b−D2a=−2b2a=−bax_1 + x_2 = \left(\frac{-b + \sqrt{D}}{2a}\right) + \left(\frac{-b - \sqrt{D}}{2a}\right) = \frac{-b + \sqrt{D} - b - \sqrt{D}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}. This is also known as Vieta's formula for the sum of the roots.

Question 8

What is the sum of all real solutions to the equation x4−8x2+5=0x^4 - 8x^2 + 5 = 0?

  1. 0 (correct answer)
  2. 8
  3. 222\sqrt{2}
  4. 8\sqrt{8}
Explanation: This equation is quadratic in form. Let u=x2u = x^2. The equation becomes u2−8u+5=0u^2 - 8u + 5 = 0. The solutions for uu are u=8±64−202=4±11u = \frac{8 \pm \sqrt{64 - 20}}{2} = 4 \pm \sqrt{11}. Both 4+114 + \sqrt{11} and 4−114 - \sqrt{11} are positive, so there are four real solutions for xx: x=±4+11x = \pm\sqrt{4 + \sqrt{11}} and x=±4−11x = \pm\sqrt{4 - \sqrt{11}}. For every positive solution, there is a corresponding negative solution. The sum of these four solutions will be (4+11)+(−4+11)+(4−11)+(−4−11)=0(\sqrt{4 + \sqrt{11}}) + (-\sqrt{4 + \sqrt{11}}) + (\sqrt{4 - \sqrt{11}}) + (-\sqrt{4 - \sqrt{11}}) = 0.

Question 9

If ii is the imaginary unit, which of the following is a solution to x2−6x+11=0x^2 - 6x + 11 = 0?

  1. 6+i26 + i\sqrt{2}
  2. 6−i26 - i\sqrt{2}
  3. 3+i23 + i\sqrt{2} (correct answer)
  4. 3+2i23 + 2i\sqrt{2}
Explanation: Using the quadratic formula with a=1,b=−6,c=11a=1, b=-6, c=11: x=−(−6)±(−6)2−4(1)(11)2(1)=6±36−442=6±−82x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(11)}}{2(1)} = \frac{6 \pm \sqrt{36 - 44}}{2} = \frac{6 \pm \sqrt{-8}}{2}. The square root of -8 is −1⋅4⋅2=2i2\sqrt{-1 \cdot 4 \cdot 2} = 2i\sqrt{2}. So, x=6±2i22x = \frac{6 \pm 2i\sqrt{2}}{2}. Dividing both terms in the numerator by 2 gives x=3±i2x = 3 \pm i\sqrt{2}. The two solutions are 3+i23 + i\sqrt{2} and 3−i23 - i\sqrt{2}. Choice C is one of these solutions.

Question 10

The solutions to a quadratic equation are given by the expression x=−8±k6x = \frac{-8 \pm \sqrt{k}}{6}. If the original equation is of the form ax2+bx+c=0ax^2 + bx + c = 0 with integer coefficients, and the discriminant is 40, what is the value of kk?

  1. 10
  2. 40 (correct answer)
  3. 120
  4. 160
Explanation: The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. The term under the radical, b2−4acb^2-4ac, is the discriminant. The question states the discriminant is 40. In the given expression, the term under the radical is kk. Therefore, kk represents the discriminant, so k=40k=40. The other information can be used to verify: −b=−8⇒b=8-b=-8 \Rightarrow b=8 and 2a=6⇒a=32a=6 \Rightarrow a=3. The discriminant is b2−4ac=82−4(3)c=64−12c=40b^2-4ac = 8^2 - 4(3)c = 64-12c = 40, which means 12c=2412c=24 and c=2c=2. The equation is 3x2+8x+2=03x^2+8x+2=0, which has a discriminant of 40.

Question 11

The area of a rectangular garden is 40 square meters. The length of the garden is 3 meters more than twice its width.

Based on the passage, what is the width of the garden in meters?

  1. −4-4
  2. −3+3294\frac{-3 + \sqrt{329}}{4} (correct answer)
  3. 3+3294\frac{3 + \sqrt{329}}{4}
  4. −3+3114\frac{-3 + \sqrt{311}}{4}
Explanation: Let ww be the width of the garden. The length ll is 2w+32w + 3. The area is l×w=40l \times w = 40. Substituting the expression for ll, we get (2w+3)w=40(2w + 3)w = 40. This expands to 2w2+3w=402w^2 + 3w = 40, or 2w2+3w−40=02w^2 + 3w - 40 = 0. Using the quadratic formula with a=2,b=3,c=−40a=2, b=3, c=-40: w=−3±32−4(2)(−40)2(2)=−3±9+3204=−3±3294w = \frac{-3 \pm \sqrt{3^2 - 4(2)(-40)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 320}}{4} = \frac{-3 \pm \sqrt{329}}{4}. Since the width must be a positive value, we take the positive root: w=−3+3294w = \frac{-3 + \sqrt{329}}{4}.

Question 12

For which of the following quadratic equations are the solutions two distinct, real, and irrational numbers?

  1. x2−8x+16=0x^2 - 8x + 16 = 0
  2. 2x2+5x−3=02x^2 + 5x - 3 = 0
  3. 3x2−4x+2=03x^2 - 4x + 2 = 0
  4. x2+5x−1=0x^2 + 5x - 1 = 0 (correct answer)
Explanation: The nature of the solutions is determined by the discriminant, D=b2−4acD = b^2 - 4ac. For two distinct, real, and irrational solutions, DD must be positive and not a perfect square. A) D=(−8)2−4(1)(16)=64−64=0D = (-8)^2 - 4(1)(16) = 64 - 64 = 0 (one real rational solution). B) D=52−4(2)(−3)=25+24=49=72D = 5^2 - 4(2)(-3) = 25 + 24 = 49 = 7^2 (two real rational solutions). C) D=(−4)2−4(3)(2)=16−24=−8D = (-4)^2 - 4(3)(2) = 16 - 24 = -8 (two non-real solutions). D) D=52−4(1)(−1)=25+4=29D = 5^2 - 4(1)(-1) = 25 + 4 = 29. Since 29 is positive and not a perfect square, this equation has two distinct, real, and irrational solutions.

Question 13

A projectile is launched from ground level, and its height hh (in feet) after tt seconds is given by h=−16t2+64th = -16t^2 + 64t. If the projectile must be caught at a height of 48 feet, at what time(s) can it be caught?

  1. t=1t = 1 second only
  2. t=3t = 3 seconds only
  3. t=1t = 1 second or t=3t = 3 seconds (correct answer)
  4. t=2t = 2 seconds or t=4t = 4 seconds
Explanation: We need to solve −16t2+64t=48-16t^2 + 64t = 48. Rearranging: −16t2+64t−48=0-16t^2 + 64t - 48 = 0. Dividing by -16: t2−4t+3=0t^2 - 4t + 3 = 0. Using the quadratic formula: t=4±16−122=4±22t = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}. This gives t=3t = 3 or t=1t = 1. Both solutions are valid since the projectile reaches 48 feet twice: once on the way up and once on the way down.

Question 14

The parabola y=3x2+7x−2y = 3x^2 + 7x - 2 intersects the x-axis at two points. What is the x-coordinate of the point that is farther from the origin?

  1. −7−736\frac{-7 - \sqrt{73}}{6} (correct answer)
  2. 7+736\frac{7 + \sqrt{73}}{6}
  3. −7+736\frac{-7 + \sqrt{73}}{6}
  4. 7−736\frac{7 - \sqrt{73}}{6}
Explanation: The x-intercepts are found when y=0y=0, so we must solve 3x2+7x−2=03x^2 + 7x - 2 = 0. Using the quadratic formula: x=−7±72−4(3)(−2)2(3)=−7±49+246=−7±736x = \frac{-7 \pm \sqrt{7^2 - 4(3)(-2)}}{2(3)} = \frac{-7 \pm \sqrt{49 + 24}}{6} = \frac{-7 \pm \sqrt{73}}{6}. The two solutions are x1=−7+736x_1 = \frac{-7 + \sqrt{73}}{6} and x2=−7−736x_2 = \frac{-7 - \sqrt{73}}{6}. To find which is farther from the origin, we compare their absolute values. Since 73\sqrt{73} is between 8 and 9 (approx 8.5), x1≈−7+8.56≈1.56=0.25x_1 \approx \frac{-7 + 8.5}{6} \approx \frac{1.5}{6} = 0.25. And x2≈−7−8.56≈−15.56≈−2.58x_2 \approx \frac{-7 - 8.5}{6} \approx \frac{-15.5}{6} \approx -2.58. The absolute value of x2x_2 is larger, so −7−736\frac{-7 - \sqrt{73}}{6} is farther from the origin.

Question 15

For the equation x2−2tx+t2−9=0x^2 - 2tx + t^2 - 9 = 0, what are the solutions for xx in terms of tt?

  1. t±3t \pm 3 (correct answer)
  2. −t±3-t \pm 3
  3. t±9t \pm 9
  4. 2t±32t \pm 3
Explanation: This equation can be solved using the quadratic formula where a=1a=1, b=−2tb=-2t, and c=t2−9c = t^2-9. x=−(−2t)±(−2t)2−4(1)(t2−9)2(1)=2t±4t2−4t2+362=2t±362=2t±62x = \frac{-(-2t) \pm \sqrt{(-2t)^2 - 4(1)(t^2-9)}}{2(1)} = \frac{2t \pm \sqrt{4t^2 - 4t^2 + 36}}{2} = \frac{2t \pm \sqrt{36}}{2} = \frac{2t \pm 6}{2}. Dividing the terms in the numerator by 2 gives x=t±3x = t \pm 3.

Question 16

If one of the solutions to the equation 2x2+bx−5=02x^2 + bx - 5 = 0 is x=12x = \frac{1}{2}, what is the other solution?

  1. 55
  2. −52-\frac{5}{2}
  3. 52\frac{5}{2}
  4. −5-5 (correct answer)
Explanation: For a quadratic equation ax2+bx+c=0ax^2+bx+c=0, the product of the two solutions is ca\frac{c}{a}. In this equation, a=2a=2 and c=−5c=-5, so the product of the solutions is x1⋅x2=−52x_1 \cdot x_2 = \frac{-5}{2}. Given that one solution is x1=12x_1 = \frac{1}{2}, we can find the other: 12⋅x2=−52\frac{1}{2} \cdot x_2 = -\frac{5}{2}. Solving for x2x_2: x2=−52÷12=−52⋅21=−5x_2 = -\frac{5}{2} \div \frac{1}{2} = -\frac{5}{2} \cdot \frac{2}{1} = -5.

Question 17

The equation 4x2−8x+1=04x^2 - 8x + 1 = 0 has solutions x1x_1 and x2x_2. Which of the following expressions represents these solutions?

  1. 2±34\frac{2 \pm \sqrt{3}}{4}
  2. 2±32 \pm \sqrt{3}
  3. 1±31 \pm \sqrt{3}
  4. 1±321 \pm \frac{\sqrt{3}}{2} (correct answer)
Explanation: Use the quadratic formula with a=4,b=−8,c=1a=4, b=-8, c=1. x=−(−8)±(−8)2−4(4)(1)2(4)=8±64−168=8±488x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(1)}}{2(4)} = \frac{8 \pm \sqrt{64 - 16}}{8} = \frac{8 \pm \sqrt{48}}{8}. Simplify the radical: 48=16⋅3=43\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}. So, x=8±438x = \frac{8 \pm 4\sqrt{3}}{8}. This fraction can be split into 88±438=1±32\frac{8}{8} \pm \frac{4\sqrt{3}}{8} = 1 \pm \frac{\sqrt{3}}{2}.

Question 18

The sum of the solutions to the equation kx2−10x+5=0kx^2 - 10x + 5 = 0 is 2. What is the product of the solutions?

  1. 12\frac{1}{2}
  2. 1 (correct answer)
  3. 52\frac{5}{2}
  4. 5
Explanation: The sum of the solutions for a quadratic equation ax2+bx+c=0ax^2+bx+c=0 is −ba-\frac{b}{a}. Here, a=ka=k and b=−10b=-10. We are given that the sum is 2. So, −−10k=2⇒10k=2⇒k=5-\frac{-10}{k} = 2 \Rightarrow \frac{10}{k} = 2 \Rightarrow k=5. Now we know the equation is 5x2−10x+5=05x^2 - 10x + 5 = 0. The product of the solutions is ca\frac{c}{a}. Here, c=5c=5 and a=5a=5. The product is 55=1\frac{5}{5} = 1.

Question 19

Which of the following represents the solutions to x+4x=4x + \frac{4}{x} = 4?

  1. x=2x = 2 (correct answer)
  2. x=2±22x = 2 \pm 2\sqrt{2}
  3. x=4,x=1x = 4, x = 1
  4. x=4x = 4
Explanation: First, multiply every term by xx to clear the fraction (assuming x≠0x \neq 0): x2+4=4xx^2 + 4 = 4x. Rearrange into standard quadratic form: x2−4x+4=0x^2 - 4x + 4 = 0. This is a perfect square trinomial, (x−2)2=0(x-2)^2 = 0, so the only solution is x=2x=2. Alternatively, using the quadratic formula with a=1,b=−4,c=4a=1, b=-4, c=4: x=−(−4)±(−4)2−4(1)(4)2(1)=4±16−162=4±02=2x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)} = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2. There is exactly one real solution.

Question 20

If x=−3+172x = \frac{-3 + \sqrt{17}}{2} is one solution to the equation x2+3x+c=0x^2 + 3x + c = 0, what is the value of cc?

  1. c=−2c = -2 because the product of roots equals c1\frac{c}{1} (correct answer)
  2. c=−1c = -1 because the product of roots equals c1\frac{c}{1}
  3. c=1c = 1 because the product of roots equals c1\frac{c}{1}
  4. c=2c = 2 because the product of roots equals c1\frac{c}{1}
Explanation: Using the quadratic formula for x2+3x+c=0x^2 + 3x + c = 0, the solutions are x=−3±9−4c2x = \frac{-3 \pm \sqrt{9 - 4c}}{2}. Since one solution is −3+172\frac{-3 + \sqrt{17}}{2}, we have 9−4c=17\sqrt{9 - 4c} = \sqrt{17}. Squaring both sides: 9−4c=179 - 4c = 17, so −4c=8-4c = 8, giving c=−2c = -2.