Solving Quadratic Equations — Print
Four methods, one goal. Factoring is fastest when it works; the quadratic formula always works.
What the practice 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.
A vegetable bed has an area of 40 m² and is 3 m longer than it is wide. How wide is it?
Call the width w. The length is w + 3, so the area is w(w + 3) = 40.
Multiply it out and something new appears: w2 + 3w = 40. There is a squared term, and the balance-scale moves cannot remove it.
Get everything to one side — w2 + 3w - 40 = 0 — and factor: (w + 8)(w - 5) = 0.
A product is zero only if one of its factors is zero. So w = -8 or w = 5.
Two answers, and only one is a garden. A quadratic often hands you two solutions and leaves you to reject one — the algebra does not know that widths cannot be negative.
Get it to zero first
Every method needs the equation in the form ax2 + bx + c = 0. Move everything to one side before starting.
x2 + 5x = 6 must become x2 + 5x - 6 = 0. Factoring the left side of the first version proves nothing.
The zero product property only works against zero. Against 6 it says nothing whatsoever.
Never divide by x
For x2 = 5x, dividing by x gives x = 5 and loses the solution x = 0. Factor instead: x(x - 5) = 0, so x = 0 or x = 5.
Factoring and square roots
If it factors, use the zero product property: set each factor to zero and solve.
If there is no x term, take square roots directly — and remember both signs. x2 = 49 gives x = ± 7.
(x - 3)2 = 16 gives x - 3 = ± 4, so x = 7 or x = -1.
Plus or minus
A square root step always produces two answers. Writing only the positive one throws away half the solution.
The quadratic formula
x = -b ± √b2 - 4ac2a works on every quadratic, factorable or not.
Write down a, b and c with their signs first. Then substitute with brackets.
Compute the discriminant on its own line before doing anything else. It keeps the arithmetic manageable.
Discriminant on its own line
Work out b2 - 4ac separately and write it down. Then put that single number under the root. Doing it all at once inside the formula is where sign errors breed.
What the discriminant tells you
b2 - 4ac > 0: two different real solutions, and the parabola crosses the x-axis twice.
b2 - 4ac = 0: one repeated solution, and the vertex sits exactly on the x-axis.
b2 - 4ac < 0: no real solutions, and the parabola misses the x-axis entirely.
A perfect square means it factors
If the discriminant is a perfect square, the roots are rational and the quadratic would have factored. Useful to know before you commit to a method.
Completing the square
Halve b, square it, and add it to both sides. x2 + 6x = 7 becomes x2 + 6x + 9 = 16.
The left side is now (x + 3)2, so x + 3 = ± 4 and x = 1 or x = -7.
This is how the quadratic formula is derived, and it is the method that converts standard form to vertex form.
Worked examples
By factoring
Solve x2 + 5x = 6.
- Move everything to one side: x2 + 5x - 6 = 0 The zero product property needs a zero.
- Two numbers multiplying to -6, adding to 5: 6 and -1. Opposite signs, since the product is negative.
- (x + 6)(x - 1) = 0 Factored.
- x = -6 or x = 1 Check x = 1: 1 + 5 = 6 ✓
x = -6 or x = 1
Quadratic formula
Solve 2x2 - 7x - 3 = 0, leaving the answer exact.
- a = 2, b = -7, c = -3 Signs included.
- Discriminant: (-7)2 - 4(2)(-3) = 49 + 24 = 73 On its own line. Positive, so two real roots.
- x = 7 ± √734 -b = 7 and 2a = 4.
- 73 is not a perfect square, so this is the exact answer. Approximately 3.89 and -0.39.
x = 7 ± √734
Completing the square
Solve x2 - 8x + 3 = 0 by completing the square.
- Move the constant: x2 - 8x = -3 Get the variable terms alone.
- Half of -8 is -4; squared is 16. Add to both sides: x2 - 8x + 16 = 13 Making a perfect square.
- (x - 4)2 = 13 The left side is now a square.
- x - 4 = ±√13, so x = 4 ± √13 Both signs. About 7.61 and 0.39.
x = 4 ± √13
Words to know
- 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.
Mistakes to watch for
Factoring before moving everything to one side. The zero product property needs a zero. From (x - 4)(x + 1) = 6 you can conclude nothing — 6 has many factor pairs.
Dividing both sides by x. From x2 = 5x that gives x = 5 and silently throws away x = 0. Factor to x(x - 5) = 0 and both roots survive.
Keeping only the positive square root. (x - 2)2 = 25 means x - 2 = ± 5, so x = 7 or x = -3. A square root step always produces two answers.
Mis-signing -4ac when c is already negative. With c = -3, -4ac becomes +24 and the discriminant grows. Compute b2 - 4ac on its own line before it goes anywhere near the formula.
Babylonian clay tablets from around 1800 BC solve quadratics by a procedure equivalent to completing the square.
They had no negative numbers, so they simply never reported the negative root — it was not that they missed it; it was not a number to them.
The formula in its modern form, with both roots and a ±, took until the 1600s and required Europe to accept negatives as numbers at all.
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