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1 Quadratic Equations; Solution by Factoring

# 1 Quadratic Equations; Solution by Factoring

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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)).

3. Set each factor equal to zero.

4. Solve the resulting linear equations. These numbers are the roots of the

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.

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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

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.

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x2 + 4 = 4x

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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

224

CHAPTER 7

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

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 6.00-V 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 first-stage 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 round-trip 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

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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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

x - 3 = { 217

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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

x-term and square this result, we have the number that completes the square. In our

example, the numerical coefficient of the x-term was 6, and 9 was added to complete

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 one-half 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 one-half 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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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

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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

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

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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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 =

- 1-2v0 2 { 2 1-2v02 2 - 4g1-22 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,

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 x-intercepts 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

232

\$)"15&3

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 flat-screen 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 3.80-m by 5.00-m rectangular

room, the area would be increased by 11.0 m2. How much must

56. Two pipes together drain a wastewater-holding 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?

"OTXFSUP1SBDUJDF&YFSDJTF

A ϭ 262 m 2

w

1. x =

-1 { 261

6

Fig. 7.7

7.4

The Graph of the Quadratic Function

5IF1BSBCPMB t 7FSUFYBOEyJOUFSDFQU t

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 ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

1 Quadratic Equations; Solution by Factoring

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