Solving Quadratic Equations — Practice
Four methods, one goal. Factoring is fastest when it works; the quadratic formula always works.
What these questions cover
These questions use 10 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Area — Space inside, measured in square units.
- Perfect square — A number that is some whole number squared: 1, 4, 9, 16, 25, …
- Square root — For k ≥ 0, √k is the nonnegative number whose square is k.
- Factor — A number that divides evenly into another.
- Parabola — The U-shaped graph of a quadratic function.
- Roots — The x values where y = 0.
- Zero product property — If AB = 0 then A = 0 or B = 0.
- Quadratic formula — x = -b ± √b2 - 4ac2a.
- Discriminant — b2 - 4ac. Its sign counts the real solutions.
- Completing the square — Rewriting x2 + bx as a perfect square minus a constant.
Problem 1
Solve x2 - 7x + 12 = 0.
Worked answer
3 and 4
(x-3)(x-4) = 0
x = 3 or x = 4
Problem 2
Solve x2 + 5x - 14 = 0.
Worked answer
-7 and 2
(x + 7)(x - 2) = 0
x = -7 or x = 2
Problem 3
Solve x2 = 81.
Worked answer
9 and -9
x = ± 9
Problem 4
Solve (x - 2)2 = 25.
Worked answer
7 and -3
x - 2 = ± 5
x = 7 or x = -3
Problem 5
Solve x2 = 6x.
Worked answer
0 and 6
x2 - 6x = 0
x(x - 6) = 0
x = 0 or x = 6
Dividing by x would have lost 0
Problem 6
Solve x2 + 4x - 5 = 0 using the quadratic formula.
Worked answer
1 and -5
b2 - 4ac = 16 + 20 = 36
x = -4 ± 62
x = 1 or x = -5
Problem 7
Solve 2x2 - 5x - 3 = 0.
Worked answer
3 and -1/2
25 + 24 = 49
x = 5 ± 74
x = 3 or x = -12
Problem 8
Find the discriminant of x2 - 6x + 9 = 0 and say how many real solutions there are.
Worked answer
0, one solution
36 - 36 = 0
Zero means one repeated solution
x = 3
Problem 9
How many real solutions does x2 + 2x + 5 = 0 have?
Worked answer
none
4 - 20 = -16
Negative, so no real solutions
The parabola misses the x-axis
Problem 10
Solve x2 + 6x = 16 by completing the square.
Worked answer
2 and -8
x2 + 6x + 9 = 25
(x+3)2 = 25
x + 3 = ± 5
x = 2 or x = -8
Problem 11
Given (x - 4)(x + 1) = 6, which step is valid?
- Expand and move everything to one side
- Set x - 4 = 6
- Set x + 1 = 6
- Set both factors to 6
Worked answer
Expand and move everything to one side
You cannot split a product that equals 6
Expand first: x2 - 3x - 10 = 0
Problem 12
Solve x2 - 4x + 1 = 0, leaving the answer exact.
Worked answer
2 + sqrt(3) and 2 - sqrt(3)
16 - 4 = 12
x = 4 ± √122 = 4 ± 2√32
x = 2 ± √3
Problem 13
Solve 3x2 + 2x - 4 = 0, leaving the answer exact.
Worked answer
(-1 + sqrt(13))/3 and (-1 - sqrt(13))/3
4 + 48 = 52
x = -2 ± √526 = -2 ± 2√136
x = -1 ± √133
Problem 14
A rectangle is 3 cm longer than it is wide and has area 40 cm2. Find its width.
Worked answer
5
w(w + 3) = 40
w2 + 3w - 40 = 0
(w + 8)(w - 5) = 0
w = 5 cm, since a width cannot be -8
Problem 15
A ball is thrown so h(t) = -16t2 + 48t. When does it hit the ground?
Worked answer
3
-16t(t - 3) = 0
t = 0 (thrown) or t = 3
3 seconds
Problem 16
For what value of k does x2 + kx + 9 = 0 have exactly one solution, with k > 0?
Worked answer
6
k2 - 36 = 0
k2 = 36
k = 6 since k > 0