1 Quadratic Equations; Solution by Factoring
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Quadratic Equations
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Recall that the solution of an equation consists of all numbers (roots) which, when
substituted in the equation, give equality. There are two roots for a quadratic equation.
At times, these roots are equal (see Example 3), so only one number is actually a solution. Also, the roots can be imaginary, and if this happens, all we wish to do at this
point is to recognize that they are imaginary.
E X A M P L E 3 Solutions (roots) of a quadratic equation
(a) The quadratic equation 3x2  7x + 2 = 0 has roots x = 1>3 and x = 2. This is
seen by substituting these numbers in the equation.
3 1 13 2 2  7 1 13 2 + 2 = 3 1 19 2 
7
3
+ 2 =
1
3

7
3
+ 2 =
0
3
= 0
2
■ Until the 1600s, most mathematicians did not
accept negative, irrational, or imaginary roots
of an equation. It was also generally accepted
that an equation had only one root.
3122  7122 + 2 = 3142  14 + 2 = 14  14 = 0
(b) The quadratic equation 4x2  4x + 1 = 0 has a double root (both roots are the
same) of x = 1>2. Showing that this number is a solution, we have
4 1 12 2 2  4 1 12 2 + 1 = 4 1 14 2  2 + 1 = 1  2 + 1 = 0
(c) The quadratic equation x2 + 9 = 0 has the imaginary roots x = 3j and x = 3j,
which means x = 32 1 and x = 32 1.
■
LEARNING TIP
The multiplicity of a root refers to
how many times a particular root
forms a solution of an equation. For
double roots of quadratics, they are
said to have a multiplicity of 2.
This section deals only with quadratic equations whose quadratic expression is factorable. Therefore, all roots will be rational. Using the fact that
a product is zero if any of its factors is zero
we have the following steps in solving a quadratic equation.
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1. Collect all terms on the left and simplify (to the form of Eq. (7.1)).
2. Factor the quadratic expression.
3. Set each factor equal to zero.
4. Solve the resulting linear equations. These numbers are the roots of the
quadratic equation.
5. Check the solutions in the original equation.
LEARNING TIP
Quadratics are often easier to graph
and factor if a 7 0 and the coefficients
of the equation are integers. This can
be accomplished by multiplying the
entire equation by  1 if necessary to
ensure a 7 0 and/or by multiplying the
entire equation by the LCD if there are
fractional coefficients.
1SBDUJDF&YFSDJTF
2
1. Solve for x: x + 4x  21 = 0
E X A M P L E 4 Solving a quadratic equation by factoring
x2  x  12 = 0
1x  42 1x + 32 = 0
x  4 = 0
x + 3 = 0
x = 3
x = 4
factor
set each factor equal to zero
solve
The roots are x = 4 and x = 3. We can check them in the original equation by substitution. Therefore,
?
?
142 2  142  12 = 0
1 32 2  1 32  12 = 0
0 = 0
Both roots satisfy the original equation.
0 = 0
■
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7.1 Quadratic Equations; Solution by Factoring
223
E X A M P L E 5 Solving a quadratic equation by factoring with a 3 1
2N 2 + 7N  4 = 0
12N  12 1N + 42 = 0
2N  1 = 0,
N =
N + 4 = 0,
1
2
factor
set each factor to zero and solve
N = 4
1
2
Therefore, the roots are N = and N = 4. These roots can be checked by the same
procedure used in Example 4.
■
E X A M P L E 6 3FBSSBOHJOHBOFRVBUJPOJOUPHFOFSBMRVBESBUJDGPSN
x2 + 4 = 4x
1SBDUJDF&YFSDJTF
x  4x + 4 = 0
2. Solve for x: 9x2 + 1 = 6x
When solving a quadratic equation, it
is important to write the equation in
the general form as a first step, setting
the quadratic equation to be equal to
zero. If the product is equal to zero,
then setting each factor equal to zero
will lead to the roots. If the product is
not zero, then it is likely that neither
factor will give a correct root.
subtract 4x from both sides
2
1x  22 = 0
x  2 = 0,
LEARNING TIP
equation not in form of Eq. (7.1)
2
factor
x = 2
solve
Since 1x  22 2 = 1x  22 1x  22, both factors are the same. This means there
is a double root of x = 2. Substitution shows that x = 2 satisfies the original
equation.
■
A number of equations involving fractions lead to quadratic equations after the fractions are eliminated. The following two examples, the second being a stated problem,
illustrate the process of solving such equations with fractions.
E X A M P L E 7 Fractional equation solved as a quadratic
Solve for x:
1
2
+ 3 =
x
x + 2
Note first that x ≠ 0, 2. We rearrange the fractions so there is a common denominator. This can be achieved by placing every term in the equation over x1 x + 22,
the LCD.
11x + 22 + 3x1x + 22
2x
=
x1x + 22
x1x + 22
Since the denominators are equal, the numerators must also be equal
1x + 22 + 3x1x + 22 = 2x
Alternatively, we could have started by multiplying each term of the original equation
by the LCD.
x1x + 22
2x1x + 22
+ 3x1x + 22 =
x
x + 2
x + 2 + 3x2 + 6x = 2x
reduce each term
2
3x + 5x + 2 = 0
13x + 22 1x + 12 = 0
3x + 2 = 0,
x + 1 = 0,
multiply each term by the LCD,
x 1x + 2 2
collect terms on left
factor
2
3
x = 1
x = 
set each factor equal
to zero and solve
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224
CHAPTER 7
Quadratic Equations
Checking in the original equation, we have
1
 23
2
?
+ 3 =
 23
1
2
?
+ 3 =
1
1 + 2
+ 2
3
? 2
? 2
+ 3 = 4
1 + 3 =
2
1
3
3
6
=
2 = 2
2
4
3
3
=
2
2
We see that the roots check. Remember, if either value gives division by zero, the root
is extraneous, and must be excluded from the solution.
■

E X A M P L E 8 Solving a word problem with quadratics
A lumber truck travels 60.0 km from a sawmill to a lumber camp and then back in 7.00 h
travel time. If the truck averages 5.00 km>h less on the return trip than on the trip to the
camp, find its average speed to the camp. See Fig. 7.1.
Let v = the average speed (in km>h) of the truck going to the camp. This means
that the average speed of the return trip was 1v  52 km>h.
We also know that d = vt (distance equals speed times time), which tells us that
t = d>v. Thus, the time for each part of the trip is the distance divided by the speed.
v km/h
60 km
Total travel
time 7 h
time to
camp
time from
camp
total
time
60
60
+
= 7
v
v  5
601v  52 + 60v = 7v1v  52
7v 2  155v + 300 = 0
v − 5 km/h
17v  152 1v  202 = 0
multiply each term by v 1v  5 2
collect terms on the left
factor
7v  15 = 0,
v =
15
7
set each factor equal to zero and solve
v  20 = 0
v = 20
The value v = 15>7 km>h cannot be a solution since the return speed of 5.00 km>h
less would be negative. Thus, the solution is v = 20.0 km>h, which means the return
speed was 15.0 km>h. The trip to the camp took 3.00 h, and the return took 4.00 h,
which shows the solution checks.
■
Fig. 7.1
E XE R C I SES 7 .1
In Exercises 1 and 2, make the given changes in the indicated examples
of this section and then solve the resulting quadratic equations.
1. In Example 5, change the + sign before 7N to  and then solve.
2. In Example 7, change the numerator of the first term to 2 and the
numerator of the term on the right to 1 and then solve.
In Exercises 3–8, determine whether or not the given equations are
quadratic. If the resulting form is quadratic, identify a, b, and c, with
a 7 0. Otherwise, explain why the resulting form is not quadratic.
3. x1x  22 = 4
2
5. x = 1x + 22
2
7. n1n2 + n  12 = n3
4. 13x  22 2 = 2
2
6. x12x + 52 = 7 + 2x
In Exercises 9–38, solve the given quadratic equations by factoring.
9. x2  4 = 0
2
11. 4x = 9
12. x2 = 0.16
13. x2  8x  9 = 0
14. x2 + x  6 = 0
2
15. R + 12 = 7R
16. x2 + 30 = 11
17. 40x  16x2 = 0
18. 15L = 20L2
2
19. 27m = 3
20. a2x2 = 9
21. 3x2  13x + 4 = 0
22. A2 + 8A + 16 = 0
2
2
8. 1T  72 2 = 12T + 32 2
10. B2  400 = 0
23. 7x + 3x = 4
24. 4x2 + 25 = 20x
25. 6x2 = 13x  6
26. 6z2 = 6 + 5x
27. 4x1x + 12 = 3
28. 9t 2 = 9  t143 + t2
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7.2 Completing the Square
30. 2x2  7ax + 4a2 = a2
29. 6y2 + by = 2b2
2
2
31. 8s + 16s = 90
32. 18t = 48t  32
33. 1x + 22 3 = x3 + 8
34. V 1V 2  42 = V 2 1V  12
35. 1x + a2 2  b2 = 0
2
2
37. x + 2ax = b  a
2
36. 2x2 = 2b2  3xb
38. x2 1a2 + 2ab + b2 2 = x1a + b2
39. In Eq. (7.1), for a = 2, b =  7, and c = 3, show that the sum of
the roots is  b>a.
40. For the equation of Exercise 39, show that the product of the roots
is c>a.
41. The voltage V across a semiconductor in a computer is given by
V = aI + bI 2, where I is the current (in A). If a 6.00V battery
is conducted across the semiconductor, find the current if
a = 2.00 Ω and b = 0.500 Ω>A.
42. The mass m (in Mg) of the fuel supply in the firststage booster of
a rocket is m = 135  6t  t 2, where t is the time (in s) after
launch. When does the booster run out of fuel? (Round to 3 significant digits.)
43. The power P (in MW) produced between midnight and noon by a
nuclear power plant is P = 4h2  48h + 744, where h is the
hour of the day. At what time is the power 664 MW?
In Exercises 51–54, set up the appropriate quadratic equations and solve.
51. The spring constant k is the force
F divided by the amount x the
spring stretches 1k = F>x2. See
Fig. 7.2(a). For two springs in
series (see Fig. 7.2(b)), the reciprocal of the spring constant kc for
the combination equals the sum of
the reciprocals of the individual
spring constants. Find the spring
constants for each of two springs
in series if kc = 2 N>cm and one
spring constant is 3 N>cm more
than the other.
In Exercises 47–50, solve the given equations involving fractions.
47.
1
4
+ = 2
x
x  3
48. 2 
49.
1
3
1
 =
2x
4
2x + 3
50.
1
3
=
x
x + 2
x
1
+
= 3
2
x  3
k ϩ3
F
(a)
(b)
Fig. 7.2
R1
R2
(a)
R2
(b)
Fig. 7.3
53. A hydrofoil made the roundtrip of 120 km between two islands
in 3.5 h of travel time. If the average speed going was 10 km>h
less than the average speed returning, find these speeds.
54. A rectangular solar panel is 20 cm by 30 cm. By adding the same
amount to each dimension, the area is doubled. How much is added?
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1. x = 7, x = 3
7.2
}x
R1
In Exercises 45 and 46, although the equations are not quadratic,
factoring will lead to one quadratic factor and the solution can be
completed by factoring as with a quadratic equation. Find the three
roots of each equation.
46. x3  4x2  x + 4 = 0
k
52. The reciprocal of the combined resistance R of two resistances R1
and R2 connected in parallel (see Fig. 7.3(a)) is equal to the sum
of the reciprocals of the individual resistances. If the two resistances are connected in series (see Fig. 7.3(b)), their combined
resistance is the sum of their individual resistances. If two resistances connected in parallel have a combined resistance of 3.0 Ω
and the same two resistances have a combined resistance of 16 Ω
when connected in series, what are the resistances?
44. In determining the speed v (in km>h) of a car while studying its
fuel economy, the equation v 2  16v = 3072 is used. Find v to
the nearest km/h.
45. x3  x = 0
225
2. x = 1>3, x = 1>3
Completing the Square
Solving a Quadratic Equation by
Completing the Square
Most quadratic equations that arise in applications cannot be solved by factoring.
Therefore, we now develop a method, called completing the square, that can be used
to solve any quadratic equation. In the next section, this method is used to develop a
general formula that can be used to solve any quadratic equation.
In the first example, we show the solution of a type of quadratic equation that arises
while using the method of completing the square. In the examples that follow it, the
method itself is used and described.
E X A M P L E 1 Quadratic equation solutions
■ The method of completing the square is
used later in Section 7.4 and also in Chapters 21,
28, and 31.
In solving x2 = 16, we may write x2  16 = 0, and complete the solution by factoring as a difference of squares. This yields 1x + 42 1x  42 = 0, giving solutions
x = 4 and x = 4.
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Quadratic Equations
We can also solve the original equation by taking the square root of each side of the
expression. Since x2 = 16, we have x = { 216, and thus x = {4. Therefore the
principal root of 16 and its negative both satisfy the original equation.
We may solve 1x  32 2 = 16 in a similar fashion:
1x  32 2 = 16
x  3 = {4
x = 1 or x = 7
Even if the number on the right side is not a perfect square, we can still use the same
method for equations of the same form. For example, we solve 1x  32 2 = 17 as
1x  32 2 = 17
x  3 = { 217
x = 3 + 217 or x = 3  217
Decimal approximations of these roots are 7.12 and 1.12.
■
E X A M P L E 2 Method of completing the square
To find the roots of the quadratic equation
x2  6x  8 = 0
first note that the left side is not factorable. However,
x2  6x is part of the special product 1x  32 2 = x2  6x + 9
and this special product is a perfect square. By adding 9 to x2  6x, we have 1x  32 2.
Therefore, we rewrite the original equation as
x2  6x = 8
x2  6x + 9 = 17
1x  32 2 = 17
add 9 to both sides
x  3 = { 217
1SBDUJDF&YFSDJTF
1. Solve by completing the square:
x2 + 4x  12 = 0
rewriting the left side
as in third illustration of Example 1
The { sign means that x  3 = { 217 or x  3 =  217.
By adding 3 to each side, we obtain
x = 3 { 217
which means x = 3 + 217 and x = 3  217 are the two roots of the equation.
Therefore, by creating an expression that is a perfect square and then using the principal square root and its negative, we were finally able to solve the equation as two
linear equations.
■
How we determine the number to be added to complete the square is based on the
special products in Eqs. (6.3) and (6.4). We rewrite these as
1x + a2 2 = x2 + 2ax + a2
1x  a2 2 = x2  2ax + a2
(7.2)
(7.3)
We must be certain that the coefficient of the x2 term is 1 before we start to complete the square. The coefficient of x in each case is numerically 2a, and the number added to complete the square is a2. Thus, if we take half the coefficient of the
xterm and square this result, we have the number that completes the square. In our
example, the numerical coefficient of the xterm was 6, and 9 was added to complete
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7.2 Completing the Square
227
the square. Therefore, the procedure for solving a quadratic equation by completing
the square is as follows:
Solving a Quadratic Equation by Completing the Square
1. Divide each side by a (the coefficient of x2).
2. Rewrite the equation with the constant on the right side.
3. Complete the square: Add the square of onehalf of the coefficient of x to
both sides.
4. Write the left side as a square and simplify the right side.
5. Equate the square root of the left side to the principal square root of the right
side and to its negative.
6. Solve the two resulting linear equations.
E X A M P L E 3 Method of completing the square
Solve 2x2 + 16x  9 = 0 by completing the square.
9
= 0
2
9
x2 + 8x =
2
x2 + 8x 
1
182 = 4;
2
1. Divide each term by 2 to make
coefficient of x2 equal to 1.
2. Put constant on right by adding
9>2 to each side.
42 = 16
9
41
+ 16 =
2
2
41
1x + 42 2 =
2
41
x + 4 = {
A2
41
x = 4 {
A2
x2 + 8x + 16 =
■ Quadratic equations can be solved using the
Solver feature of some graphing calculators.
3. Divide coefficient 8 of x by 2,
square the 4, and add to
both sides.
4. Write left side as 1x + 4 2 2.
5. Take the square root of each side
and equate the x + 4 to the
principal square root of 41>2 and
its negative.
6. Solve for x.
41
Therefore, the roots are 4 + 241
2 and 4  2 2 .
■
The method of completing the square is useful in situations other than the solutions
of quadratic equations. For example, writing an expression as the sum of two squares is
an important algebraic step when computing some integrals (Section 28.6) and when
finding inverse Laplace transforms (Section 31.11). Example 4 illustrates how the
square is completed for such cases.
E X A M P L E 4 Writing an expression as the sum of two squares
Write the expression x2 + 4x + 13 as the sum of two squares.
The square of onehalf of the coefficient of x is 4, so 4 is the number that completes
the square. We rewrite 13 as the sum 13 = 4 + 9, so that
x2 + 4x + 13 = x2 + 4x + 4 + 9
= (x + 2)2 + 9
= (x + 2)2 + 32
■
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Quadratic Equations
& 9& 3 $ * 4 & 4
In Exercises 1 and 2, make the given changes in the indicated
examples of this section and then solve the resulting quadratic
equations by completing the square.
1. In Example 2, change the  sign before 6x to + .
2. In Example 3, change the coefficient of the second term from 16
to 12.
In Exercises 3–10, solve the given quadratic equations by finding
appropriate square roots as in Example 1.
3. x2 = 25
4. x2 = 100
2
2
5. x = 7
6. x = 15
2
7. 1x  22 = 25
2
9. 1x + 32 = 7
8. 1x + 22 2 = 10
10. 1 x 
2
5 2
2
= 100
In Exercises 11–30, solve the given quadratic equations by completing
the square. Exercises 11–14 and 17–20 may be checked by factoring.
11. x2 + 2x  8 = 0
12. x2  8x  20 = 0
13. D2 + 3D + 2 = 0
14. t 2 + 5t  6 = 0
2
15. n = 4n  2
17. v1v + 22 = 15
2
16. 1R + 92 1R + 12 = 13
25. 5T 2  10T + 4 = 0
26. 4V 2 + 9 = 12V
27. 9x + 6x + 1 = 0
28. 2x2  3x + 2a = 0
29. x2 + 2bx + c = 0
30. px2 + qx + r = 0
2
In Exercises 31 and 32, use completing the square to write the given
expression as the sum of two squares.
31. x2 + 6x + 13
32. x2  8x + 17
In Exercises 33–36, use completing the square to solve the given problems.
33. The voltage V across a certain electronic device is related to the
temperature T (in °C) by V = 4.00T  0.200T 2. For what
temperature(s) is V = 15.0 V ?
34. A flare is shot vertically into the air such that its distance s (in m)
above the ground is given by s = 20.0t  5.00t 2, where t is the
time (in s) after it was fired. Find t for s = 15.0 m.
35. A surveillance camera is 12.0 m on a direct line from an ATM.
The camera is 5.00 m more to the right of the ATM than it is
above the ATM. How far above the ATM is the camera?
36. A rectangular storage area is 8.00 m longer than it is wide. If the
area is 28.0 m2, what are its dimensions?
18. Z 2 + 12 = 8Z
19. 2s + 5s = 3
20. 4x2 + x = 3
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21. 3y2 = 3y + 2
22. 3x2 = 3  4x
23. 2y2  y  2 = 0
24. 9v 2  6v  2 = 0
1. x = 6, x = 2
7.3
The Quadratic Formula
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We now use the method of completing the square to derive a general formula that may
be used for the solution of any quadratic equation.
Consider Eq. (7.1), the general quadratic equation:
ax2 + bx + c = 0
When we divide through by a, we obtain
x2 +
1a ≠ 02
b
c
x + = 0
a
a
Subtracting c>a from each side, we have
x2 +
b
c
x = a
a
Half of b>a is b>2a, which squared is b2 >4a2. Adding b2 >4a2 to each side gives us
x2 +
b
b2
c
b2
x +
=  +
2
a
a
4a
4a2
Writing the left side as a perfect square and combining fractions on the right side,
we have
2
b
b2  4ac
ax +
b =
2a
4a2
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7.3 The Quadratic Formula
Equating x +
b
to the principal square root of the right side and its negative,
2a
x +
■ The quadratic formula, with the { sign,
means that the solutions to the quadratic
equation
are
229
b
{ 2b2  4ac
=
2a
2a
When we subtract b>2a from each side and simplify the resulting expression, we obtain
the quadratic formula:
ax 2 + bx + c = 0
x =
b + 2b 2  4ac
2a
x =
b  2b 2  4ac
2a
x =
b { 2b2  4ac
2a
(7.4)
and
The quadratic formula gives us a quick general way of solving any quadratic equation. We need only write the equation in the standard form of Eq. (7.1), substitute these
numbers into the formula, and simplify.
E X A M P L E 1 Quadratic formula—rational roots
Solve: x2

a = 1
5x
+
6
b = 5
0.
=
c = 6
Here, using the indicated values of a, b, and c in the quadratic formula, we have
 1 52 { 2 1 52 2  4112 162
5 { 225  24
5 { 1
=
=
2112
2
2
5 + 1
5  1
x =
= 3 or x =
= 2
2
2
x =
The roots x = 3 and x = 2 check when substituted in the original equation.
COMMON ERROR
■
It must be emphasized that, in using the quadratic formula, the entire expression
 b { 2b2  4ac is divided by 2a. It is a relatively common error to divide only the radical
2b2  4ac by 2a.
E X A M P L E 2 Quadratic formula—irrational roots
Solve: 2x2
a = 2
Practice Exercise
1. Solve using the quadratic formula:
3x2 + x  5 = 0

7x
b = 7

5
= 0.
c = 5
Substituting the values for a, b, and c in the quadratic formula, we have
 1 72 { 2 1 72 2  4122 1 52
7 { 249 + 40
7 { 289
=
=
2122
4
4
7 + 289
7  289
x =
= 4.108 or x =
= 0.6085
4
4
x =
7 { 289
(this form is often used when the roots are irrational).
4
Approximate decimal values are x = 4.11 and x = 0.608.
■
The exact roots are x =
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Quadratic Equations
E X A M P L E 3 Quadratic formula—double root
Solve: 9x2 + 24x + 16 = 0.
In this example, a = 9, b = 24, and c = 16. Thus,
x =
24 { 2242  4192 1162
24 { 2576  576
24 { 0
4
=
=
= 2192
18
18
3
Here, both roots are  43, and we write the result as x =  43 and x =  43. We will get a
double root when b2 = 4ac, as in this case.
■
E X A M P L E 4 Quadratic formula—imaginary roots
Solve: 3x2  5x + 4 = 0.
In this example, a = 3, b = 5, and c = 4. Therefore,
x =
 1 52 { 2 1 52 2  4132 142
5 { 225  48
5 { 2 23
=
=
2132
6
6
These roots contain imaginary numbers. This happens if b2 6 4ac.
■
Examples 1–4 illustrate the character of the roots of a quadratic equation. If a, b,
and c are rational numbers (see Section 1.1), by noting the value of b2  4ac (called
the discriminant), we have the following:
LEARNING TIP
If b2  4ac is positive and a perfect
square, ax 2 + bx + c is factorable. We
can use the value of b2  4ac to help
in checking the roots or in finding
the character of the roots without
having to solve the equation
completely.
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1. If b2  4ac is positive and a perfect square (see Section 1.6), the roots are
real, rational, and unequal. (See Example 1, where b2  4ac = 1.)
2. If b2  4ac is positive but not a perfect square, the roots are real, irrational,
and unequal. (See Example 2, where b2  4ac = 89.)
3. If b2  4ac = 0, the roots are real, rational, and equal. (See Example 3,
where b2  4ac = 0.)
4. If b2  4ac 6 0, the roots contain imaginary numbers and are unequal. (See
Example 4, where b2  4ac = 23.)
E X A M P L E 5 Quadratic formula—literal numbers
g is the
acceleration
due to gravity
t is the time
s
Fig. 7.4
1
2
gt 2 is used in the analysis of projectile motion (see
gt 2  2v0t  21s0  s 2 = 0
v0
s0
The equation s = s0 + v0 t Fig. 7.4). Solve for t.
multiply by  2, put in form of Eq. (7.1)
In this form, we see that a = g, b = 2v0, and c = 21s0  s 2 :
t =
t =
 12v0 2 { 2 12v02 2  4g122 1s0  s 2
2v0 { 241v 20 + 2gs0  2gs 2
=
2g
2g
2v0 { 22v 20 + 2gs0  2gs
v0 { 2v 20 + 2gs0  2gs
=
g
2g
■
E X A M P L E 6 Quadratic formula—word problem
A rectangular area 17.0 m long and 12.0 m wide is to be used for a patio with a rectangular pool. One end and one side of the patio area around the pool (the chairs, sunning,
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7.3 The Quadratic Formula
231
etc.) are to be the same width. The other end with the diving board is to be twice as
wide, and the other side is to be three times as wide as the narrow side. The pool area is
to be 96.5 m2. What are the widths of the patio ends and sides, and the dimensions of
the pool? See Fig. 7.5.
First, let x = the width of the narrow end and side of the patio. The other end is then
2x in width, and the other side is 3x in width. Since the pool area is 96.5 m2, we have
2x
pool length
pool width
pool area
117.0  3x2 112.0  4x2 = 96.5
Patio
204  68.0x  36.0x + 12x2 = 96.5
12x2  104.0x + 107.5 = 0
Pool
17.0 m
x =
96.5 m2
x
 1 104.02 { 2 1 104.02 2  41122 1107.52
104.0 { 25656
=
21122
24
Evaluating, we get x = 7.47 m and x = 1.20 m. The value 7.47 m cannot be the
required result since the width of the patio would be greater than the width of the entire
area. For x = 1.20 m, the pool would have a length 13.4 m and width 7.20 m. These
give an area of 96.5 m2, which checks. The widths of the patio area are then 1.20 m,
1.20 m, 2.40 m, and 3.60 m.
■
3x
x
12.0 m
Fig. 7.5
& 9& 3$* 4 & 4
In Exercises 1– 4, make the given changes in the indicated examples
of this section and then solve the resulting equations by the quadratic
formula.
1. In Example 1, change the  sign before 5x to + .
2. In Example 2, change the coefficient of x2 from 2 to 3.
3. In Example 3, change the + sign before 24x to  .
4. In Example 4, change 4 to 3.
In Exercises 5–36, solve the given quadratic equations, using the
quadratic formula. Exercises 5–8 are the same as Exercises 11–14 of
Section 7.2.
5. x2 + 2x  8 = 0
2
7. D + 3D + 2 = 0
2
9. x  4x + 2 = 0
6. x2  8x  20 = 0
2
8. t + 5t  6 = 0
10. x2 + 10x  4 = 0
11. v 2 = 15  2v
12. 8V  12 = V 2
13. 2s2 + 5s = 3
14. 4x2 + x = 3
15. 3y2 = 3y + 2
16. 3x2 = 3  4x
17. y + 2 = 2y2
18. 2 + 6v = 9v 2
2
19. 30y + 23y  40 = 0
2
2
20. 40x  62x  63 = 0
21. 8t + 61t =  120
22. 2d1d  22 = 7
23. s2 = 9 + s 11  2s 2
24. 20r 2 = 20r + 1
29. x2  0.200x  0.400 = 0
30. 3.20x2 = 2.50x + 7.60
25. 25y2 = 121
27. 15 + 4z = 32z
2
2
31. 0.290Z  0.180 = 0.630Z
26. 37T = T 2
28. 4x2  12x = 7
2
32. 12.5x + 13.2x = 15.5
33. x2 + 2cx  1 = 0
35. b2x2 + 1  a = 1b + 12x
34. x2  7x + 16 + a2 = 0
36. c2x2  x  1 = x2
In Exercises 37– 40, without solving the given equations, determine
the character of the roots.
37. 2x2  7x = 8
2
39. 3.6t + 2.1 = 7.7t
38. 3x2 + 19x = 14
40. 0.45s2 + 0.33 = 0.12s
In Exercises 41–58, solve the given problems. All numbers are
accurate to at least 3 significant digits.
41. Find k if the equation x2 + 4x + k = 0 has a real double root.
42. Find the smallest positive integral value of k if the equation
x2 + 3x + k = 0 has roots with imaginary numbers.
43. Solve the equation x4  5x2 + 4 = 0 for x. (Hint: The equation
can be written as 1x2 2 2  51x2 2 + 4 = 0. First solve for x2.)
44. Without drawing the graph or completely solving the equation,
explain how to find the number of xintercepts of a quadratic
function.
45. In machine design, in finding the outside diameter D0 of a hollow
shaft, the equation D20  DD0  0.250D2 = 0 is used. Solve for
D0 if D = 3.625 cm.
46. A missile is fired vertically into the air. The distance s (in m)
above the ground as a function of time t (in s) is given by
s = 100 + 500t  4.9t 2. (a) When will the missile hit the
ground? (b) When will the missile be 1000 m above the ground?
47. For a rectangle, if the ratio of the length to the width equals the
ratio of the length plus the width to the length, the ratio is called
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232
$)"15&3
Quadratic Equations
the golden ratio. Find the value of the golden ratio, which the
ancient Greeks thought had the most pleasing properties to look at.
48. When focusing a camera, the distance r the lens must move from
the infinity setting is given by r = f 2 > 1p  f2, where p is the
distance from the object to the lens, and f is the focal length of
the lens. Solve for f .
54. Two circular oil spills are tangent to each other. If the distance
between centres is 800 m and they cover a combined area of
1.02 * 106 m2, what is the radius of each? See Fig. 7.8.
1.02 × 106 m2
49. In calculating the current in an electric circuit with an inductance
L, a resistance R, and a capacitance C, it is necessary to solve the
equation Lm2 + Rm + 1>C = 0. Solve for m in terms of L, R,
and C. See Fig. 7.6.
800 m
Fig. 7.8
L
R
C
Fig. 7.6
50. In finding the radius r of a circular arch of height h and span b, an
architect used the following formula. Solve for h.
b2 + 4h2
r =
8h
51. A flatscreen computer monitor is 37.0 cm wide and 31.0 cm high
with a uniform edge around the viewing screen. If the edge covers
20.0% of the monitor front, what is the width of the edge?
52. An investment of $2000 is deposited at a certain annual interest
rate. One year later, $3000 is deposited in another account at the
same rate. At the end of the second year, the accounts have a total
value of $5319.05. What is the interest rate?
53. The length of a tennis court is 12.8 m more than its width. If the area
of the tennis court is 262 m2, what are its dimensions? See Fig. 7.7.
w ϩ 12.8 m
55. In remodeling a house, an architect finds that by adding the same
amount to each dimension of a 3.80m by 5.00m rectangular
room, the area would be increased by 11.0 m2. How much must
be added to each dimension?
56. Two pipes together drain a wastewaterholding tank in 6.00 h.
If used alone to empty the tank, one takes 2.00 h longer than
the other. How long does each take to empty the tank if used
alone?
57. On some highways, a car can legally travel 20.0 km>h faster than
a truck. Travelling at maximum legal speeds, a car can travel 120 km
in 18.0 min less than a truck. What are the maximum legal speeds
for cars and for trucks?
58. For electric capacitors connected in series, the sum of the
reciprocals of the capacitances equals the reciprocal of the
combined capacitance. If one capacitor has 5.00 mF more
capacitance than another capacitor and they are connected in
series, what are their capacitances if their combined capacitance is 4.00 mF?
"OTXFSUP1SBDUJDF&YFSDJTF
A ϭ 262 m 2
w
1. x =
1 { 261
6
Fig. 7.7
7.4
The Graph of the Quadratic Function
5IF1BSBCPMB t 7FSUFYBOEyJOUFSDFQU t
Solving Quadratic Equations Graphically
y
x
y
24
5
23
0
22
23
21
24
0
23
1
0
2
5
In this section, we discuss the graph of the quadratic function ax2 + bx + c and show
the graphical solution of a quadratic equation. By letting y = ax2 + bx + c, we can
graph this function, as in Chapter 3. The next example shows the graph of the quadratic function.
4
E X A M P L E 1 Graphing a quadratic function
2
24
22
0
2
x
Graph the function f1x2 = x2 + 2x  3.
First, let y = x2 + 2x  3. We can now set up a table of values and graph the function as shown in Fig. 7.9.
■
22
24
Fig. 7.9
The graph of the quadratic function shown in Fig. 7.9 is that of a parabola, and the
graph of any quadratic function y = ax2 + bx + c will have the same basic shape. (In
Section 3.4, we briefly noted that graphs in Examples 2 and 3 were parabolas.) A