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BROCK UNIVERSITY MATHEMATICS MODULES
11A2.3: Completing The Square
Author: Kristina Wamboldt
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• What it is: Completing the square is a way of changing the form of a quadratic expression
to make it easier to work with in some situations.
• Why you need it: We complete the square to help us solve a quadratic equation, or to
help us determine the vertex of the parabolic graph of a quadratic function.
• When to use it: Completing the square will help you determine the vertex of a parabola.
(Once you know calculus you will learn a faster method.) Completing the square will
also help you to describe transformations of parabolas, which can help you in graphing.
PREREQUISITES
Before you tackle this module, make sure you have completed these modules:
Expanding products of binomials, Factoring quadratic expressions, Zeros of
quadratic functions, Graphing quadratic functions
WARMUP
Before you tackle this module, make sure you can solve the following exercises. If you have
difficulties, please review the appropriate prerequisite modules.
(Answers below.)
1. Expand the following.
(b) (x + 4)2 (c) (x − 5)2
!
"
!
"
1 2
2 2
(e) x +
(f) 5 x −
2
3
(a) (x − 2)(x + 6)
(d) 2(x − 2)2
2. Factor the following.
(a) x2 − 100
(b) x2 + 18x + 81
(d) 3x2 + 24x + 48
(c) 25x2 − 120x + 144
(e) 18x2 − 12x + 2
(f) 4x2 − 12x + 9
1
3. (a) Sketch a graph of y = x2 .
(b) What are the coördinates of the vertex of the graph in Part (a)?
4. Determine the x-intercepts and y-intercepts of the graph of each function.
(a) y = x2 − 9
(b) y = −3x2 + 6
3
(c) y = − x2 + 3x
2
5. Sketch a graph of each function.
(a) y = 2(x − 1)2 − 3
(b) y = −(x + 2)2 + 1
Answers: 1.(a) x2 + 4x − 12 (b) x2 + 8x + 16 (c) x2 − 10x + 25 (d) 2x2 − 8x + 8 (e) x2 + x + 14 (f)
5x2 − 20
x+ 20
2.(a) (x+10)(x−10) (b) (x+9)2 (c) (5x−12)2 (d) 3(x+4)2 (e) 2(3x−1)2 = 18(x−1/3)2 (f)
3
9
√
√
2
(2x − 3) = 4(x − 3/2)2 3.(b) (0, 0) 4.(a) (3, 0), (−3, 0), (0, −9) (b) ( 2, 0), (− 2, 0), (0, 6) (c) (2, 0), (0, 0)
5.
(a)
(b)
Introduction
Completing the square is a technique that’s used to convert the formula for a quadratic function
from standard form to what’s known as vertex form. This allows us to determine some key features
of the graph of a quadratic function (which is a parabola), such as the coordinates of the vertex
and the equation of the axis of symmetry. It’s also helpful in solving quadratic equations that can’t
easily be placed in factored form; in fact, the completing the square technique is used in Module
*** to derive the quadratic formula, which enables us to solve any quadratic equation.
In this module you’ll learn how to use the technique of completing the square, and apply it to
some problems.
Completing the square can be quite difficult at first, but keep practicing and you’ll soon be able
to answer questions such as the following.
FOCUS QUESTION
To help you understand an important aspect of this lesson, focus your attention on this
question, which will be answered towards the end of the lesson.
Juliet stands on the balcony of her third floor apartment. She leans out and throws
a ball straight up with an initial speed of 20 m/s. As you no doubt expected,
Romeo is standing on the ground, waiting to catch the ball. The ball’s height y
in metres above the ground t seconds after it is thrown can be approximated by
the formula y = −5t2 + 20t + 11.25.
2
(a) When does the ball reach its peak height?
(b) What is the ball’s peak height?
(c) Romeo is overcome by his love for Juliet and fails to notice the ball as it
reaches him, so it falls at his feet without him catching it. When does the
ball hit the ground?
The goal of this module is to learn how to write expressions such as 2x2 + 12x + 13, which is
said to be in standard form, in the form 2(x + 3)2 − 5, which is said to be in vertex form. For some
purposes standard form is easier to work with, but for other purposes vertex form is easier to work
with, so it is good to be able to move from one form to the other if necessary.
First let’s consider converting a quadratic expression from vertex form to standard form. Study
the following steps:
2(x + 3)2 − 5 = 2(x + 3)(x + 3) − 5
= 2(x2 + 3x + 3x + 9) − 5
= 2(x2 + 6x + 9) − 5
= 2x2 + 12x + 18 − 5
= 2x2 + 12x + 13
The vertex form of the quadratic expression in the first line has been converted, step-by-step, to
the standard form in the last line. The two forms are equivalent, in the sense that their values are
identical for each x-value, but their forms are different.
The idea behind completing the square is to run this process backwards; that is, start with the
standard form in the last line and proceed step-by-step until we reach the vertex form in the first
line. But it seems difficult to outline a procedure that will work in all cases; for instance, how on
earth would we know to write 13 as 18 − 5 in going from the last line to the one before it?
INVESTIGATION
Explore this on your own!
It’s worthwhile trying to figure out the procedure of completing the square on your own, even
if only for ten or fifteen minutes. The effort of doing so, even if unsuccessful,a prepares your
mind to understand the procedure, and helps you to internalize it and remember it. Try this
frequently when you are learning mathematics, and you’ll be pleasantly surprised!
a
If you are successful, even partly successful, it’s a tremendous bonus.
After playing around for a while with this example, you might hit upon the key idea: When
you expand (x + 3)(x + 3) you end up with x2 + 6x + 9. The key is to notice that the 3 is doubled
to produce 6, but squared to produce 9. In going backwards to produce the vertex form, we would
take half of the 6 to produce the 3, and then make sure that there is a 9 separated from whatever
the constant term. Notice that 9 is the square of 3; another way to say this is that 9 is the square
of half of 6.
3
Let’s see how this works in detail for this example; here’s how to convert the standard form
2x2 + 12x + 13 to the vertex form 2(x + 3)2 − 5 in a step-by-step way:
2x2 + 12x + 13 = 2(x2 + 6x) + 13
(factor 2 from the first two terms)
= 2(x2 + 6x + 9 − 9) + 13
(add and subtract the square of half of 6)
= 2(x2 + 6x + 9) − 5
(combine the constant terms)
2
= 2(x + 6x + 9) − 18 + 13 (note: 2(9) = −18; move the −18 term outside the parentheses)
2
= 2(x + 3) − 5
(factor the expression in parentheses)
KEY IDEA
Here are the steps for completing the square of a quadratic expression of the form ax2 + bx + c:
1. Factor the coefficient of x2 from the first two terms. If the coefficient of x2 is 1, just enclose
the first two terms with parentheses.
2. Take the current coefficient of x (that is, the one inside the parentheses) and divide it by
2. Then square the result.
3. Add and subtract the final result of the previous step inside the parentheses; note that this
is the same as adding 0 inside the parentheses, and therefore is a valid algebraic operation.
4. Move the negative constant factor outside the parentheses, multiplying it by the factor in
front of the parentheses first.
5. Combine the constant terms outside the parentheses.
6. Factor the expression in the parentheses to write it as a square.
EXAMPLE 1
Complete the square for the following quadratic expression. y = x2 + 8x − 2
SOLUTION
1. Place parentheses around the first two terms.
y = (x2 + 8x) − 2
2. Divide the coefficient of the x-term by 2 and square the result.
8
=4
2
42 = 16
4
3. Take the final result of Step 2, and then add it and subtract it inside the parentheses. (In
effect, this adds 0 to the expression, which therefore does not change its value.)
y = (x2 + 8x + 16 − 16) − 2
4. Remove the negative term from the parentheses. (If there is a factor in front of the parentheses, make sure to multiply the negative term by it as the term is removed.)
y = (x2 + 8x + 16) − 16 − 2
5. Simplify by combining the terms outside the parentheses.
y = (x2 + 8x + 16) − 18
6. Factor the expression within the parentheses and write it as a square.
y = (x + 4)2 − 18
Follow these 6 easy steps to complete the square for any quadratic expression! Here’s another
example:
EXAMPLE 2
Complete the square for the following quadratic expression.
y = x2 − 2x − 8
SOLUTION
Step 1:
y = (x2 − 2x) − 8
Step 2:
2
=1
2
=⇒
12 = 1
Step 3:
y = (x2 − 2x + 1 − 1) − 8
Step 4:
y = (x2 − 2x + 1) − 1 − 8
Step 5:
y = (x2 − 2x + 1) − 9
Step 6:
y = (x − 1)2 − 9
5
Now it’s your turn to try a few!
PRACTICE
(Answers below.)
1. Complete the square for each quadratic expression.
(a) y = x2 + 10x + 8
(b) y = x2 + 3x − 8
Answers: 1.(a) y = (x + 5)2 − 17 (b) y = (x + 3/2)2 − 41/4
Quadratic expressions for which the coefficient of x2 is not 1 are a bit more challenging; let’s
look at one next:
EXAMPLE 3
Complete the square for the quadratic expression.
y = 2x2 + 8x + 5
SOLUTION
With this example, notice that there is a 2 in front of the x2 . The first step is to factor 2 from
the first two terms:
y = 2(x2 + 4x) + 5
Continuing the step-by-step process, we obtain:
Step 2:
4
=2
2
=⇒
22 = 4
Step 3:
y = 2(x2 + 4x + 4 − 4) + 5
Step 4:
y = 2(x2 + 4x + 4) − 8 + 5
Notice that as we removed the −4 from the parentheses, we had to multiply it by 2, the factor
at the front of the parentheses.
Step 5:
y = 2(x2 + 4x + 4) − 3
Step 6:
y = 2(x + 2)2 − 3
6
Let’s now look at an example that is a bit more complicated.
EXAMPLE 4
Complete the square for the quadratic expression.
y = −4x2 + 7x − 3
SOLUTION
As in the previous examples, we’ll just go through the 6 steps.
Step 1:
!
"
7
y = −4 x − x − 3
4
2
Notice that the coefficient of x is a fraction. This is still fine; we’re still able to work with this
and finish completing the square. Let’s continue the step-by-step procedure.
Step 2:
Step 3:
! "
7
7
4
=
2
8
! "2
7
49
=
8
64
=⇒
!
2
!
2
7
49 49
y = −4 x − x +
−
4
64 64
Step 4:
Step 6:
−3
"
+
49
−3
16
7
49
y = −4 x − x +
4
64
"
+
7
49
y = −4 x − x +
4
64
Step 5:
"
!
2
!
7
y = −4 x −
8
"2
+
1
16
1
16
As you can see, even if the coefficient of x is a fraction, we can still complete the square. Just
follow the 6 steps and all will work well.
PRACTICE
(Answers below.)
Complete the square for each quadratic expression.
2. y = 5x2 + 5x − 2
3. y = 3x2 − 6x + 2
7
4. y = −3x2 − 9x + 5
5. y = −8x2 + 15x − 2
3
6. y = x2 − x + 4
2
1
7. y = x2 + 2x − 6
4
Answers: 2. y = 5(x + 1/2)2 − 13/4 3. y = 3(x − 1)2 − 1 4. y = −3(x + 3/2)2 + 47/4 5. y = −8(x − 1)2 + 6
6. y = (3/2)(x − 1/3)2 + 23/6 7. y = (1/4)(x + 4)2 − 10
Now that you understand how to place a quadratic expression into vertex form, let’s look at
how you can determine the vertex of a parabola using vertex form:
EXAMPLE 5
Find the vertex of the parabola that is the graph of the quadratic function.
y = −4x2 + 7x − 3
SOLUTION
As you can see, this is the same quadratic function that appeared in Example 4. As we determined
in Example 4, the vertex form of the function is
!
7
y = −4 x −
8
"2
+
1
16
Recall from Module *** that when the formula for a quadratic function is placed in vertex form,
the coördinates of the vertex can be read from the formula, as shown below:
y = a(x − h)2 + k
(h, k) is the vertex
The coördinates of the vertex are (h, k). Using this for our example, we get:
8
!
7
y = −4 x −
8
!
7 1
,
8 16
"2
"
+
1
16
is the vertex
As you can see, the vertex of the graph of the function y =
−4x2
+ 7x − 3 is
!
"
7 1
,
.
8 16
Now that we know how to determine the vertex of the graph of a quadratic function, let’s take
a look at the focus question.
RECAP OF FOCUS QUESTION
Recall the focus question, which was asked earlier in the lesson.
Juliet stands on the balcony of her third floor apartment. She leans out and throws
a ball straight up with an initial speed of 20 m/s. As you no doubt expected,
Romeo is standing on the ground, waiting to catch the ball. The ball’s height y
in metres above the ground t seconds after it is thrown can be approximated by
the formula y = −5t2 + 20t + 11.25.
(a) When does the ball reach its maximum height?
(b) What is the ball’s maximum height?
(c) Romeo is overcome by his love for Juliet and fails to notice the ball as it
reaches him, so it falls at his feet without him catching it. When does the
ball hit the ground?
SOLUTION
Parts (a) and (b) amount to calculating the coördinates of the vertex of the graph of the
quadratic position function. So, let’s complete the square:
y = −5t2 + 20t + 11.25
= −5(t2 − 4t) + 11.25
= −5(t2 − 4t + 4 − 4) + 11.25
= −5(t2 − 4t + 4) + 20 + 11.25
= −5(t2 − 4t + 4) + 31.25
= −5(t − 2)2 + 31.25
9
Now that the formula for the position function is in vertex form, we can read the coördinates
of the vertex from the formula: the vertex is at (2, 31.25). This means that the ball reaches
its maximum height at a time of 2 s after it is thrown, and its maximum height is 31.25 m
above the ground.a
For Part (c), the ground is described by y = 0, so to determine when the ball reaches the
ground we’ll substitute y = 0 into the position function and solve for t. Having the position
function in vertex form will make the solution process relatively quick:
0 = −5(t − 2)2 + 31.25
5(t − 2)2 = 31.25
31.25
(t − 2)2 =
5
2
(t − 2) = 6.25
√
t − 2 = ± 6.25
√
t = 2 ± 6.25
t = 2 ± 2.5
t = −0.5 s
and
t = 4.5 s
Therefore, the ball lands at Romeo’s feet 4.5 s after Juliet throws it. The solution t = −0.5 s
is rejected because it represents a time before the ball is thrown, and is therefore not relevant
for this situation.
a
Juliet has clearly taken her vitamins today.
INVESTIGATION
Explore this on your own!
In the previous example, what about the rejected solution, t = −0.5 s? In the context of the
problem, we reject this solution, since the negative value for time indicates a time before the
ball was thrown. However, there is nothing intrinsically wrong with negative times, contrary
to what you may have heard. We use negative times to describe the BC (also known as BCE)
era of history; saying 25 BC is the same as saying −25. However, one must look carefully at
the context of the problem and reject any candidate solutions that lie outside the domain of the
problem. This is one of the reasons why the concept of the domain of a function is important.
(See Module *** for more on domain and range.)
However, sometimes rejected candidate solutions reveal hidden insights that we would never
have obtained if we simply rejected them without thinking any further. Examine the rejected
solution t = −0.5 s in the previous example. Make sense of it. What I mean by this is that you
should strive to think of a second problem, related to the one in the previous example, where
the solution does make sense. This may give you more insight into the previous example, into
quadratic equations, into the properties of parabolas, and into motion problems.
Finally, here are a couple of examples that might seem odd at first. Suppose we wish to complete
the square for each of the following functions: y = x2 + 9, and y = x2 − 6x. Both may seem a
little strange; the first is missing an x-term, and the second is missing a constant term. However,
neither of them is problematic, as you can see below:
10
1. y = x2 + 9
This equation is already in vertex form (y = (x−0)2 +9), therefore we do not need to complete
the square.
2. y = x2 − 6x
This equation is the same as y = x2 − 6x + 0, so we can proceed with the usual steps to
complete the square:
y = (x2 − 6x) + 0
6
=3
=⇒
32 = 9
2
y = (x2 − 6x + 9 − 9) + 0
y = (x2 − 6x + 9) − 9 + 0
y = (x2 − 6x + 9) − 9
y = (x − 3)2 − 9
WWW
• What we did: We learned how to convert a quadratic equation from standard form to
vertex form by completing the square.
• Why we did it: It’s helpful to be able to convert from standard form to vertex form,
because each has its advantages in different situations.
• What’s next: Next we can use vertex form to help us graph quadratic functions, and
further understand the transformations that can be applied to basic quadratic functions.
EXERCISES
1. Find the value of k that makes each trinomial a perfect square.
(a) x2 + 6x + k
(b) x2 − 22x + k
2. Complete the square for each of the following.
(a) y = x2 + 8x + 5
(b) y = x2 − 4x + 1
(c) y = 2x2 − 3x − 5
(d) y = 2x2 − 8x + 3
(e) y = 3x2 + 4x − 2
11
3. Find the coördinates of the vertex for the following.
(a) y = (x − 6)2 − 12
(b) y = (x + 7)2 + 27
(c) y = x2 + 2x + 8
(d) y = 3x2 + 6x + 1
4. Complete the following table.
Equation
y=
4x2
y=
x2
Vertex
x-intercept(s)
+ 8x − 21
+ 3x − 10
y = −3x2 + 6x − 2
12
y-intercept