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```					                                                                              Math 233 - Spring 2009
8.1      Solving Quadratic Equations by Completing the Square
8.1.1     Square Roots Revisited
Recall in the last chapter, we studied square roots and found that for any square roots there were
two possible roots:                    √            √
9 = 3,       9 = −3
At that time we deﬁned the square root to be the positive or principle square root. However, when
solving equations we want to ﬁnd all possible solutions so for example:
EX 1. Solve: x2 = 9
We square root both sides and we get: x = 3 however, we also have that x = −3 is a solution.
√
In General if x2 = a where a is a real number then x = ± a
√
REMARK 1. The notation ± is read “plus or minus” and is understood to mean that + a and
√
− a are both solutions to the equation.
EX 2. Solve:
1. t2 + 3 = 27

2. x2 + 11 = 0

3. (t + 4)2 + 27 = 0

√
REMARK 2. √ • Note that the statement t = −4±3i 3 means there are two possible solutions:
√
t = −4 + 3i 3 and t = −4 + 3i 3.
• In past chapters we solved x2 = 9 by rewriting it:
x2 − 9    =   0
(x + 3)(x − 3)   =   0
and we know that x = 3 or x = −3. However for the ﬁrst example above t2 = 24 this method
doesn’t work because t2 − 24 doesn’t factor. That will be the case for most of our polynomials
in this chapter.
• We will develop methods to solve quadratic equations that do not factor.

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8.1.2    Perfect Square Trinomials
We will learn a process called comleting the squares. In order to do this, we need to understand
perfect square trinomials. Consider:

x2 + 4x + 4    =   (x + 2)(x + 2)    =   (x + 2)2
x2 − 4x + 4    =   (x − 2)(x − 2)    =   (x − 2)2
x2 + 6x + 9    =   (x + 3)(x + 3)    =   (x + 3)2
x2 − 6x + 9    =   (x − 3)(x − 3)    =   (x − 3)2

EX 3. Let’s examine this further. Consider x2 + 6x + 9:

b
We will use this property a lot, if x2 + bx + c is a perfect square then c = ( 2 )2 .

8.1.3    Completing the Square
Steps:
1. Divide by the leading coeﬃcient. (We want the coeﬃcient of the square term to be 1).

2. Isolate the constant term on one side of the equation.
3. Take the coeﬃcient of the ﬁrst degree term, divide it by two, and square the resulting
number. This is what we add to both sides of the equation.
4. Replace the trinomial with the square of a binomial.

5. Square root both sides.
6. Solve and check.
EX 4. Solve by completing the square.

1. x2 + 10x = −21

2. r2 + 9r + 8 = 0

2
3. −x2 = −3x − 10

4. x2 − 10x + 37 = 0

5. −3a2 + 3a = −18

8.2.1   Derivation
Recall the standard form of a quadratic equation is

ax2 + bx + c = 0

where a, b, c are real numbers.
EX 5. Identify a, b, and c

1. x2 − 3x + 4

2. 2.3x2 − 7.2

3. − 2 x2 + 1 x
3      4

Let’s solve ax2 + bx + c = 0 by completing the square.

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(derivation continued)

√
−b ±    b2 − 4ac
x=
2a
This is an extremely useful formula for solving quadratic equations.

Let’s look at some examples.
EX 6. Solve using the quadratic formula.
1. x2 + 5x − 6 = 0

2. −25x2 = 10x + 1

3. m2 − 1 m +
2
3
4   =0

4. f (x) = 3x2 + 5x Find all values of x for which f (x) = −3

4
8.2.3   Working Backwards
Suppose we are given the solutions to a quadratic equation but not the equation itself. Can we
derive the equation?
EX 7. Determine an equation that has the given solutions.

1. Solutions: -3, 2

2. Solutions: 2 + i, 2 − i

REMARK 3. Why would we care about this? In an application suppose we know that our system
follows a quadratic equation. Example: Suppose we know the temperature follows a quadratic
function and we know temperature is zero at times 3 and 2

8.2.4   Number of Real Solutions
The discriminant of a quadratic equation is

b2 − 4ac

the part under the square root of the quadratic formula.
For an equation of the form
ax2 + bx + c = 0
where a = 0:
• If b2 − 4ac > 0, the equation has two distinct real number soltuions.
• If b2 − 4ac = 0, the equation has a single real number solution.

• if b2 − 4ac < 0, the equation has no real number solutions.
EX 8. -
1. For the equation x2 − 5x − 14 ﬁnd the discriminant, the number of real number solutions and

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2. For the following, determine the number of real number solutions without actually solving.
(a) 3x2 − 2x + 8 = 0

(b) 4x2 − 4x + 1 = 0

The Geometry We can visualize what is going on here if we consider the function f (x) = ax2 +bx+
c. Then our solutions (quadratic formula) are where the function f (x) = 0 (the x-intercepts). We’ve
mentioned that the graphs of quadratic functions are parabolas, we have the following graphical
possibilities:

8.3    Applications and Problem Solving
Quadratic functions appear frequently in applications. Let’s look at a couple of contrived word
problems:
EX 9. -
1. A motorboat travels 16 miles with a 6-mph current and then turns around and returns to the
starting point traveling against the current. If the entire trip takes 2 hours, ﬁnd the speed of
the motorboat in still water.

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2. Dan and Bill can paint a living room in 4 hours working together. Dan can paint the room
1 hour faster than Bill. How long will it ake each of them working alone to paint the living
room.

Deﬁnition 1. An equation that can be written in the form au2 + bu + c = 0 for a = 0, where u is
an algebraic expression, is called quadratic form.
We will be doing substitution to solve equations that are almost in quadratic form. For example:

Equation Quadratic in Form          Substition    Equation with Substitution
4     2                                    2
y −y −6=0                             u=y                 u2 − u − 6 = 0
2(x + 5)2 − 5(x + 5) − 12 = 0        u=x+5               2u2 − 5u − 12 = 0
2        1                                1
x 3 + 4x 3 − 3 = 0                    u = x3              u2 + 4u − 3 = 0

EX 10. Solve
1. y 4 − 5y 2 − 14 = 0

2. x4 − 13x2 + 36 = 0

3. x4 + 4x2 − 45 = 0

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4. 6(a + 1)2 − 7(a + 1) − 3 = 0

5. Find the x-intercepts of the graph of the function f (x) = 4x−2 − 9x−1 + 2

2     1
6. x 3 + x 3 − 12 = 0

√
7. 2m −       m − 21 = 0

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8.5.1   Characteristics
The graphs of quadratic functions are called parabolas. There are several characteristics of a parabola
that we will use to draw the graph.
f (x) = ax2 + bx + c
We have the following properties:
• Opening: A parabola can either open upwards or downwards.
– When a > 0 it opens upwards.
6

-

– When a < 0 it opens downwards.
6

-

• Axis of Symmetry: A parabola is symmetric about a given vertical line. This line is called
the axis of symmetry. The equation for the axis of symmetry is given by:
b
x=−
2a

6

-

• Vertex: The vertex is the lowest point of a parabola that opens upward or the highest point
for a parabola that opens downward.

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6                                  6

-                                     -

or
REMARK 4. The vertex is also the minimum or maximum value of the function (depending
on whether it opens up or down).

– At what value does the vertex occur?
b
– Answer: Always occur at the axis of symmetry. Thus when x = − 2a .
– Remember the vertex is a point so we should list it as a point, the formula is:
b        b
(−      , f (− ))
2a       2a
or
b 4ac − b2
(−      ,        )
2a    4a
• y-intercept: Occurs when the parabola passes through the y-axis. Occurs when x = 0. Thus
y-intercept = (0, f (0)).
• x-intercepts: Occur when the parabola passes through the x-axis. Must solve f (x) = 0.
Recall that for a parabola there can be zero, one, or two x-intercepts depending on discriminant.

– If b2 − 4ac > 0, the equation has two distinct x-intercepts.
– If b2 − 4ac = 0, the equation has a single x-intercept.
– if b2 − 4ac < 0, the equation has no x-intercepts.

Let’s now apply all of these ideas to graph some parabolas.

EX 11. Do a - e for the given equations.
a. Determine whether the parabola opens upward or downward.
b. Find the y-intercept.

c. Find the vertex
d. Find the x-intercepts if any.
e. Draw the graph.

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1. For the equation y = −x2 + 2x + 3

6

-

2. For the equation f (x) = x2 − 3x + 4

6

-

8.5.3   Maximum and Minimum Problems
Recall that the vertex gives us the maximum value if the parabola opens downard and the mini-
mum value if the parabola opens upward. For example:
y = ax2 + bx + c
For a > 0 minimum value           For a < 0 maximum value
6                                 6

-                             -

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We can use these facts to solve interesting application problems.
EX 12. -
1. Consider the following rectangle.
(a) Find an equation for the area, A(x)
(b) Find the value for x that gives the largest (maximum) area.
(c) Find the maximum area.

2. If the perimeter of a rectangle is 60 feet, ﬁnd the dimensions that will give the greatest area.

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