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ACT Math Quadratics: 22 Practice Problems with Step-by-Step Explanations

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ACT Math Quadratics: 22 Practice Problems with Step-by-Step Explanations

Quadratic equations are a cornerstone of the ACT Math section, and this chapter stretches well beyond basic factoring — you'll see quadratics disguised as area and volume word problems, quadratics built from given roots, the quadratic formula applied to equations that don't factor nicely, and even higher-degree polynomials that reduce to a quadratic through substitution. The good news is that most of these problems rely on a small set of core techniques applied in different combinations.

The essential toolkit: factoring by finding two numbers that multiply to give the constant term and add to give the middle coefficient (or factoring by grouping for more complex expressions); the quadratic formula, x = (−b ± √(b2−4ac))/(2a), for equations that resist factoring; and the sum and product of roots shortcuts — for ax2+bx+c=0, the sum of the roots is −b/a and the product of the roots is c/a — which let you skip solving entirely on certain problems. The discriminant, b2−4ac, tells you the nature of the solutions without solving: positive means two real roots, zero means one repeated real root, and negative means two complex roots.

Common traps include forgetting that a quadratic can have complex (non-real) solutions when the discriminant is negative, mixing up which sign goes with which root when writing a quadratic from its factors, and on higher-degree polynomial problems, forgetting that a repeated factor like (x+1)2 only contributes one distinct real root, not two. Work through the 22 problems below, including one that references a graph, then click to reveal each step-by-step explanation.

On this page:

  1. Problem 1 — Solving a factored equation
  2. Problem 2 — Sum of roots
  3. Problem 3 — Factoring a quadratic
  4. Problem 4 — Sum of roots
  5. Problem 5 — Area word problem
  6. Problem 6 — Product of roots
  7. Problem 7 — Quadratic formula
  8. Problem 8 — Simplifying a rational expression
  9. Problem 9 — Factoring by grouping
  10. Problem 10 — Difference of squares
  11. Problem 11 — Identifying a function from its graph
  12. Problem 12 — Roots to factors (cubic)
  13. Problem 13 — Vertex form and symmetry
  14. Problem 14 — Combining two quadratics
  15. Problem 15 — Real roots of a quartic
  16. Problem 16 — Volume word problem
  17. Problem 17 — Matching coefficients
  18. Problem 18 — Solving a quartic by substitution
  19. Problem 19 — Discriminant and root type
  20. Problem 20 — Quadratic formula
  21. Problem 21 — Quadratic formula
  22. Problem 22 — Roots to a factor
  23. Problem 22 — Y-intercepts of two functions

Practice Problems

Problem 1

What are the solutions to the quadratic equation (4x+1)(2x−3) = 0?

  • A) −1 and 3
  • B) −1/4 and −3/2
  • C) −1/4 and 3/2
  • D) 1/4 and −3/2
Correct Answer: C

Set each factor equal to 0. From 4x+1=0: x = −1/4. From 2x−3=0: x = 3/2.

Problem 2

What is the sum of the solutions of the polynomial f(x) = x2 − 7x + 12?

  • A) −7
  • B) −4
  • C) 4
  • D) 7
Correct Answer: D

For ax2+bx+c=0, the sum of the roots is −b/a. Here a=1 and b=−7, so the sum is −(−7)/1 = 7.

Problem 3

What are the roots of the quadratic equation 3x2 + 9x − 12 = 0?

  • A) x = 1 and x = 4
  • B) x = −1 and x = 4
  • C) x = −1 and x = −4
  • D) x = 1 and x = −4
Correct Answer: D

Divide the whole equation by 3: x2 + 3x − 4 = 0. Factor: find two numbers that multiply to −4 and add to 3 — those are 4 and −1: (x+4)(x−1) = 0.

So x = −4 or x = 1.

Problem 4

What is the sum of the solutions of the equation x2 − 4x − 21 = 0?

  • A) −10
  • B) −4
  • C) 4
  • D) 10
Correct Answer: C

Using the sum of roots shortcut, −b/a: here a=1, b=−4, so the sum is −(−4)/1 = 4.

Problem 5

The lengths of 2 adjacent sides of a rectangle are represented by x+4 and 3x−1 inches. What expression below represents the area of the rectangle in square inches?

  • A) 3x2 + 12x − 4
  • B) 3x2 + 11x − 4
  • C) 3x2 + 7 − 3
  • D) 3x2 − 12
Correct Answer: B

Area is the product of the two side lengths: (x+4)(3x−1). Expand using FOIL: 3x2 − x + 12x − 4.

Combine like terms: 3x2 + 11x − 4.

Problem 6

For what value of m does the quadratic x2 − 13x + m = 0 have solutions of x=5 and x=8?

  • A) −40
  • B) −13
  • C) 13
  • D) 40
Correct Answer: D

For ax2+bx+c=0, the product of the roots is c/a. Here a=1, so m equals the product of the two roots: 5 × 8 = 40.

As a check, the sum of the roots should equal −b/a = 13, and indeed 5+8=13. ✓

Problem 7

What is the solution set for 5x2 + 6x = 8?

  • A) {1/5, 1/2}
  • B) {−1/5, −1/2}
  • C) {4/5, 2}
  • D) {4/5, −2}
Correct Answer: D

Move everything to one side: 5x2 + 6x − 8 = 0. Apply the quadratic formula with a=5, b=6, c=−8: x = (−6 ± √(36+160))/10 = (−6 ± √196)/10 = (−6 ± 14)/10.

This gives x = 8/10 = 4/5 or x = −20/10 = −2.

Problem 8

For x2 ≠ 25, which of the following expressions is equal to (x−5)2 / (x2−25)?

  • A) (x−5)/(x+5)
  • B) 1/(x+5)
  • C) 1/(x−5)
  • D) 1
Correct Answer: A

Factor the denominator as a difference of squares: x2−25 = (x−5)(x+5). The expression becomes (x−5)2 / [(x−5)(x+5)].

Cancel one factor of (x−5) from the top and bottom: (x−5)/(x+5).

Problem 9

Which of the following are the factors of abx2 + (ac−bd)x − cd?

  • A) (ax−b)(cx−d)
  • B) (ax+b)(cx−d)
  • C) (ax+c)(bx−d)
  • D) (ax−d)(bx+c)
Correct Answer: D

Factor by grouping: abx2 + acx − bdx − cd = ax(bx+c) − d(bx+c).

Both terms share the factor (bx+c): (ax−d)(bx+c). Expanding this confirms it matches the original expression.

Problem 10

In the equation below, a and b are constants. Which of the following could be the value of a?

9x2 − 16 = (ax−b)(ax+b)

  • A) 3
  • B) 4
  • C) 9
  • D) 16
Correct Answer: A

The right side is a difference of squares: (ax−b)(ax+b) = a2x2 − b2. Matching this to 9x2−16 gives a2=9 and b2=16.

So a = 3 (or −3), and 3 is among the answer choices.

Problem 11

The function f(x) is graphed below. Which of the following could define the function f(x)?

A cubic function crossing the x-axis at x = −3, x = 0, and x = 1

  • A) f(x) = (x−3)(x+1)
  • B) f(x) = x(x−3)(x+1)
  • C) f(x) = (x+3)(x−1)
  • D) f(x) = x(x+3)(x−1)
Correct Answer: D

The graph crosses the x-axis at x=−3, x=0, and x=1, so f(x) must have factors of x, (x+3), and (x−1). The graph is a cubic shape (an S-curve with two turns) rather than a simple upward or downward parabola, confirming it needs all three factors, not just two.

f(x) = x(x+3)(x−1) matches all three roots and produces the correct sign pattern in each interval: negative for x<−3, positive between −3 and 0, negative between 0 and 1, and positive for x>1 — exactly matching the graph's shape.

Problem 12

In the equation below, j, l, k, and m are constants. If the equation has roots of −4, 3, and −5, which of the following could be a factor of the equation below?

jx3 + lx2 − kx − m = 0

  • A) x−4
  • B) x−5
  • C) x−3
  • D) x+3
Correct Answer: C

Each root r corresponds to a factor of (x−r). The given roots are −4, 3, and −5, so the factors are (x−(−4)) = (x+4), (x−3), and (x−(−5)) = (x+5).

Of the answer choices, only x−3 matches one of these factors.

Problem 13

In the standard (x,y) coordinate plane, the graph of the equation y = 3(x−2)2−5 intersects the x-axis at points (r,0) and (7,0) and has its vertex at point (2,−5). What is the value of r?

  • A) −7/2
  • B) −3
  • C) −5/2
  • D) −2
Correct Answer: B

A parabola is symmetric about the vertical line through its vertex. The vertex is at x=2, and one root is at x=7, which is 7−2=5 units to the right of the vertex.

The other root must be exactly 5 units to the left of the vertex: r = 2−5 = −3.

Problem 14

Given that (x+4) and (x+3) are factors of x2+cx+d and x=−3 is the only solution of the quadratic x2+mx+n, what is dm?

  • A) 18
  • B) 24
  • C) 72
  • D) 144
Correct Answer: C

First find d: since (x+4) and (x+3) are the factors, x2+cx+d = (x+4)(x+3) = x2+7x+12, so d=12.

Next find m: since x=−3 is the only (repeated) solution, x2+mx+n = (x+3)2 = x2+6x+9, so m=6.

Multiply: dm = 12 × 6 = 72.

Problem 15

h(x) = x4 + 2x3 − 8x2 − 18x − 9. The polynomial above can be written as (x2−9)(x+1)2. What are all the real roots of the equation?

  • A) 3, −3
  • B) 9, 1, and −1
  • C) 3, −3, and −1
  • D) 3, −3, 1, and −1
Correct Answer: C

Set each factor equal to 0. From x2−9=0: x=3 or x=−3. From (x+1)2=0: x=−1 (a repeated root, but still just one distinct value).

Combining all distinct real roots: 3, −3, and −1.

Problem 16

The side lengths of a box are represented by x feet, 2x−3 feet, and x+5 feet. In terms of x, what is the volume, in cubic feet, of the box?

  • A) 2x2 − 15x
  • B) 2x2 + 7x − 15
  • C) 2x3 − 15x
  • D) 2x3 + 7x2 − 15x
Correct Answer: D

Volume is the product of all three side lengths: x(2x−3)(x+5). First expand the last two factors: (2x−3)(x+5) = 2x2+10x−3x−15 = 2x2+7x−15.

Multiply by the remaining factor x: 2x3+7x2−15x.

Problem 17

Given that (2x+3) and (x−4) are the factors of the quadratic 2x2+(z−1)x+2z−4=0, what is the value of z?

  • A) −2
  • B) −4
  • C) 2
  • D) 4
Correct Answer: B

Expand the given factors: (2x+3)(x−4) = 2x2−8x+3x−12 = 2x2−5x−12.

Match coefficients with 2x2+(z−1)x+2z−4: the x coefficient gives z−1=−5, so z=−4. Checking the constant: 2z−4 = 2(−4)−4 = −12, which matches. ✓

Problem 18

What is the solution set to the equation x4 + 7x2 − 144 = 0?

  • A) {−4, −3, 3, 4}
  • B) {−4, 4, −3i, 3i}
  • C) {−3, 3, −4i, 4i}
  • D) {−16, −9}
Correct Answer: C

Substitute u = x2: u2+7u−144=0. Apply the quadratic formula: u = (−7 ± √(49+576))/2 = (−7 ± √625)/2 = (−7 ± 25)/2.

This gives u=9 or u=−16. Since u=x2: from u=9, x=±3. From u=−16, x2=−16, so x=±4i.

Full solution set: {−3, 3, −4i, 4i}.

Problem 19

Which of the following most precisely describes the solutions to the equation 3x2+5x+7=0?

  • A) 1 rational (double) root
  • B) 1 irrational (double) root
  • C) 2 rational roots
  • D) 2 complex roots (with nonzero imaginary parts)
Correct Answer: D

The discriminant, b2−4ac, tells you the nature of the roots without solving. Here a=3, b=5, c=7: discriminant = 25 − 4(3)(7) = 25−84 = −59.

Since the discriminant is negative, the equation has 2 complex roots with nonzero imaginary parts.

Problem 20

Which of the following is a solution to the equation x2+6x+3=0?

  • A) −3+√6
  • B) −3+√13
  • C) −6−2√6
  • D) 3−√6
Correct Answer: A

Apply the quadratic formula with a=1, b=6, c=3: x = (−6 ± √(36−12))/2 = (−6 ± √24)/2.

Simplify √24 = 2√6, so x = (−6 ± 2√6)/2 = −3 ± √6. One solution is −3+√6.

Problem 21

Which of the following is a solution to the equation x2−4x+1=0?

  • A) 2−√6
  • B) −2+√6
  • C) 2+√3
  • D) −2−√3
Correct Answer: C

Apply the quadratic formula with a=1, b=−4, c=1: x = (4 ± √(16−4))/2 = (4 ± √12)/2.

Simplify √12 = 2√3, so x = (4 ± 2√3)/2 = 2 ± √3. One solution is 2+√3.

Problem 22

Given that f(x) = (x−3)(x+12)+4 and g(x) = (2/7)x2 − (19/14)x − 5/7, what is the sum of the y-intercepts of the functions?

  • A) −257/7
  • B) −247/7
  • C) −229/7
  • D) −219/7
Correct Answer: C

A y-intercept is the function's value at x=0. For f: f(0) = (0−3)(0+12)+4 = (−3)(12)+4 = −36+4 = −32.

For g: g(0) = (2/7)(0)2 − (19/14)(0) − 5/7 = −5/7.

Sum: −32 + (−5/7) = −224/7 − 5/7 = −229/7.

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