ACT Math study guide · Chapter 8
Quadratic Equations
A quadratic equation can be solved by factoring, by completing the square, or with the quadratic formula. Factoring is fastest when the equation factors with integer coefficients. The quadratic formula, x = (−b ± √(b² − 4ac)) / 2a, solves any quadratic equation regardless of whether it factors.
The discriminant, b² − 4ac, determines the type of solutions a quadratic equation has before the equation is fully solved. A positive discriminant produces two real solutions. A discriminant of zero produces exactly one real solution. A negative discriminant produces two complex solutions.
The vertex of a parabola y = ax² + bx + c has x-coordinate −b/2a. Substituting that x-value into the equation gives the y-coordinate. For any quadratic ax² + bx + c = 0, the sum of the solutions equals −b/a and the product of the solutions equals c/a, without solving the equation directly.
Where this appears as a model: an object's height over time. An object launched from 5 feet with an initial upward speed of 64 feet per second follows h(t)=−16t²+64t+5; the maximum height, found at the vertex, occurs at t=2 seconds and equals 69 feet.
▸ Solving by factoring and by formula
A quadratic factors cleanly when integer coefficients exist that multiply to give ac and add to give b. When no such integers exist, or when the quadratic does not factor, the quadratic formula applies directly.
Worked Example Solve x² − 2x − 15 = 0.
Find two numbers that multiply to −15 and add to −2.
−5 and 3 satisfy both conditions.
(x − 5)(x + 3) = 0
x = 5 or x = −3
▸ The discriminant and the vertex
A negative discriminant does not stop the solving process; it signals that the two solutions will be complex conjugates, found the same way a positive discriminant is handled, with an imaginary term appearing under the radical.
Worked Example Find the discriminant of x² + 2x + 5 = 0, and state the number of real solutions.
Apply the discriminant formula directly.
(2)² − 4(1)(5) = 4 − 20 = −16
The discriminant is negative, so the equation has zero real solutions and two complex solutions.
A parameter question asks for the value of an unknown coefficient that forces a specific discriminant condition, most often exactly one real solution. Setting the discriminant equal to zero and solving for the parameter answers this question type directly.
▸ Practice
1. Solve by factoring: x² + 7x + 10 = 0.
A) x = 5, 2 B) x = −5, 2 C) x = 5, −2 D) x = −5, −2
2. Solve 2x² − 3x − 5 = 0 using the quadratic formula.
F) x = −2.5, 1 G) x = 2.5, −1 H) x = −2.5, −1 J) x = 2.5, 1
3. What is the discriminant of 3x² − 2x + 5 = 0?
A) 56; two real solutions B) −56; two complex solutions C) −56; one real solution D) 56; no solutions
4. Find the vertex of y = x² − 8x + 15.
F) (−4, −1) G) (4, 1) H) (4, −1) J) (8, 15)
5. For what value(s) of k does x² + kx + 16 = 0 have exactly one real solution?
A) k = ±8 B) k = 8 only C) k = −8 only D) k = ±4
6. The solutions to x² − 4x + 13 = 0 are complex conjugates a ± bi (b > 0). What is a + b?
F) −1 G) 7 H) 5 J) 1
7. Solve by factoring: x² − 3x − 18 = 0.
A) x = 6, −3 B) x = −6, 3 C) x = 6, 3 D) x = −6, −3
8. Solve 4x² + 4x − 3 = 0 using the quadratic formula.
F) x = −1.5, 0.5 G) x = 1.5, −0.5 H) x = −1.5, −0.5 J) x = 1.5, 0.5
9. Find the discriminant of 3x² + 2x + 5 = 0, and the number of real solutions.
A) 56; two real solutions B) −56; one real solution C) 56; no solutions D) −56; two complex solutions
10. Find the vertex of y = x² + 6x + 2.
F) (3, −7) G) (−3, 7) H) (3, 7) J) (−3, −7)
11. For what value(s) of k does x² + kx + 49 = 0 have exactly one real solution?
A) k = 14 only B) k = −14 only C) k = ±14 D) k = ±7
12. Find the sum and product of the solutions to x² − 7x + 10 = 0.
F) sum = 7, product = 10 G) sum = 10, product = 7 H) sum = −7, product = 10 J) sum = 7, product = −10
13. Find the sum and product of the solutions to x² + 4x + 29 = 0.
A) sum = −4, product = 29 B) sum = 4, product = 29 C) sum = −4, product = −29 D) sum = 4, product = −29
14. Solve by factoring: 2x² − x − 15 = 0.
F) x = −3, 2.5 G) x = 3, −2.5 H) x = 3, 2.5 J) x = −3, −2.5
15. Solve by factoring: x² − 5x − 24 = 0.
A) x = −8, 3 B) x = 8, −3 C) x = 8, 3 D) x = −8, −3
16. Solve 3x² + 5x − 2 = 0 using the quadratic formula.
F) x = 2, −1/3 G) x = −2, 1/3 H) x = −2, −1/3 J) x = 2, 1/3
17. Find the discriminant of 5x² − 4x + 1 = 0, and the number of real solutions.
A) 4; two real solutions B) −4; two complex solutions C) −4; one real solution D) 4; no solutions
18. Find the vertex of y = x² + 4x − 5.
F) (2, −9) G) (−2, 9) H) (2, 9) J) (−2, −9)
19. For what value(s) of k does x² + kx + 100 = 0 have exactly one real solution?
A) k = 20 only B) k = −20 only C) k = ±20 D) k = ±10
20. Find the sum and product of the solutions to x² + 6x + 8 = 0.
F) sum = −6, product = 8 G) sum = 6, product = 8 H) sum = −6, product = −8 J) sum = 6, product = −8
▸ Answers and Explanations
1: D. Factor: (x + 5)(x + 2) = 0, giving x = −5 or x = −2. Choice A reverses the sign of both roots.
2: G. Using the quadratic formula: discriminant = 9 + 40 = 49, √49 = 7. x = (3 ± 7)/4, giving x = 2.5 or x = −1. Choice F reverses the sign of both roots.
3: B. Discriminant: (−2)² − 4(3)(5) = 4 − 60 = −56, which is negative, giving two complex solutions. Choice A reports the discriminant with the sign reversed, along with the real-solution conclusion a positive discriminant would produce.
4: H. x-coordinate: −(−8)/2(1) = 4. y-coordinate: 4² − 8(4) + 15 = 16 − 32 + 15 = −1. Choice F reverses the sign of the x-coordinate only.
5: A. One real solution requires the discriminant to equal 0: k² − 64 = 0, so k² = 64, giving k = 8 or k = −8. Choice B reports only the positive root, dropping the negative solution.
6: H. Solving gives x = (4 ± √(16 − 52))/2 = (4 ± 6i)/2 = 2 ± 3i. With b > 0, a = 2 and b = 3, so a + b = 5. Choice F computes a − b instead of a + b.
7: A. Factor: (x − 6)(x + 3) = 0, giving x = 6 or x = −3. Choice B reverses the sign of both roots.
8: F. Using the quadratic formula: discriminant = 16 + 48 = 64, √64 = 8. x = (−4 ± 8)/8, giving x = 0.5 or x = −1.5. Choice J reverses the sign of one root only.
9: D. Discriminant: (2)² − 4(3)(5) = 4 − 60 = −56, which is negative, giving two complex solutions. Choice A reports the discriminant with the sign reversed, along with the real-solution conclusion a positive discriminant would produce.
10: J. x-coordinate: −6/2(1) = −3. y-coordinate: (−3)² + 6(−3) + 2 = 9 − 18 + 2 = −7. Choice F reverses the sign of the x-coordinate only.
11: C. One real solution requires the discriminant to equal 0: k² − 196 = 0, so k² = 196, giving k = 14 or k = −14. Choice A reports only the positive root, dropping the negative solution.
12: F. For x² − 7x + 10 = 0, the sum of the roots is −b/a = 7 and the product is c/a = 10. Choice G swaps the sum and the product.
13: A. For x² + 4x + 29 = 0, the sum of the roots is −b/a = −4 and the product is c/a = 29. Choice B reverses the sign of the sum only.
14: G. Factor: (2x + 5)(x − 3) = 0, giving x = 3 or x = −2.5. Choice F reverses the sign of both roots.
15: B. Factor: (x − 8)(x + 3) = 0, giving x = 8 or x = −3. Choice A reverses the sign of both roots.
16: G. Using the quadratic formula: discriminant = 25 + 24 = 49, √49 = 7. x = (−5 ± 7)/6, giving x = 1/3 or x = −2. Choice F reverses the sign of both roots.
17: B. Discriminant: (−4)² − 4(5)(1) = 16 − 20 = −4, which is negative, giving two complex solutions. Choice A reports the discriminant with the sign reversed, along with the real-solution conclusion a positive discriminant would produce.
18: J. x-coordinate: −4/2(1) = −2. y-coordinate: (−2)² + 4(−2) − 5 = 4 − 8 − 5 = −9. Choice F reverses the sign of the x-coordinate only.
19: C. One real solution requires the discriminant to equal 0: k² − 400 = 0, so k² = 400, giving k = 20 or k = −20. Choice A reports only the positive root, dropping the negative solution.
20: F. For x² + 6x + 8 = 0, the sum of the roots is −b/a = −6 and the product is c/a = 8. Choice G reverses the sign of the sum only.