copyright © 2010 pearson education, inc. all rights reserved sec 10.3 - 1
TRANSCRIPT
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 10.3 - 1
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 10.3 - 2
Quadratic Equations, Inequalities,
and Functions
Chapter 10
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 10.3 - 3
10.3
Equations Quadratic in Form
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 4
10.3 Equations Quadratic in Form
Objectives
1. Solve an equation with fractions by writing it in quadratic
form.
2. Use quadratic equations to solve applied problems.
3. Solve an equation with radicals by writing it in quadratic
form.
4. Solve an equation that is quadratic in form by substitution.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 5
Clear fractions by multiplying each term by the least common denominator,
6x(x – 5). (Note that the domain must be restricted to x ≠ 0 and x ≠ 5.)
Solve .3x
=2x – 5
+ 56
EXAMPLE 1 Solving an Equation with Fractions That
Leads to a Quadratic Equation
10.3 Equations Quadratic in Form
3x
=2x – 5
+ 56
6x(x – 5) 6x(x – 5) 6x(x – 5)
18(x – 5) + 12x = 5x(x – 5)
18x – 90 + 12x = 5x2 – 25x Distributive property
30x – 90 = 5x2 – 25x Combine terms.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 6
Combine and rearrange terms so that the quadratic equation is in standard
form. Then factor to solve the resulting equation.
Solve .3x
=2x – 5
+ 56
EXAMPLE 1 Solving an Equation with Fractions That
Leads to a Quadratic Equation
10.3 Equations Quadratic in Form
30x – 90 = 5x2 – 25x
5(x2 – 11x + 18) = 0 Factor.
5x2 – 55x + 90 = 0 Standard form
5(x – 2)(x – 9) = 0 Factor.
x – 2 = 0 or x – 9 = 0 Zero-factor property.
x = 2 or x = 9 Solve each equation.Check by substituting these solutions in the original equation. The solution
set is { 2, 9 }.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 7
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 1 Read the problem carefully.
Step 2 Assign a variable. Let x = the speed of the current. The current
slows down the boat when it is going upstream, so the rate (or
speed) upstream is the speed of the boat in still water less the speed
of the current, or 10 – x.
Riverboat traveling
upstream – the current
slows it down.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 8
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 2 Assign a variable. Let x = the speed of the current. The current
slows down the boat when it is going upstream, so the rate (or
speed) upstream is the speed of the boat in still water less the speed
of the current, or 10 – x.
Similarly, the current speeds up the
boat as it travels downstream, so
its speed downstream is 10 + x.
Thus,
10 – x = the rate upstream;
10 + x = the rate downstream.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 9
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 2 Assign a variable. Let x = the speed of the current. The current
slows down the boat when it is going upstream, so the rate (or
speed) upstream is the speed of the boat in still water less the speed
of the current, or 10 – x.
810 – x
810 + x
d r t
Upstream 8 10 – x
Downstream 8 10 + x
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 10
=
810 – x
810 + x
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 3 Write an equation. The total time, 1 hr and 40 min, can be written as
810 – x
810 + x
d r t
Upstream 8 10 – x
Downstream 8 10 + x
40601 + = 2
31 + = 53 hr.+
Time upstream Time downstream Total Time
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 11
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 4 Solve the equation. Multiply each side by 3(10 – x)(10 + x), the LCD,
and solve the resulting quadratic equation.
810 – x
810 + x+ = 5
3
3(10 + x)8 + 3(10 – x)8 = 5(10 – x)(10 + x)
24(10 + x) + 24(10 – x) = 5(100 – x2)
240 + 24x + 240 – 24x = 500 – 5x2 Distributive property
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 12
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 4 Solve the equation. Multiply each side by 3(10 – x)(10 + x), the LCD,
and solve resulting quadratic equation.
480 = 500 – 5x2 Combine terms.
5x2 = 20
x2 = 4 Divide by 5.
x = 2 or x = –2 Square root property
240 + 24x + 240 – 24x = 500 – 5x2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 13
A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr,
40 minutes to go 8 miles upstream and return. Find the speed of the current.
EXAMPLE 2 Solving a Motion Problem
10.3 Equations Quadratic in Form
Step 5 State the answer. The speed of the current cannot be –2, so the
answer is 2 mph.
Step 6 Check that this value satisfies the original problem.
d r t
Upstream 8 10 – x
Downstream 8 10 + x
d r t
Upstream 8 10 – 2
Downstream 8 10 + 2
d r t
Upstream 8 8
Downstream 8 12 23
1
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 14
10.3 Equations Quadratic in Form
Caution on “Solutions”
CAUTION
As shown in Example 2, when a quadratic equation is used to solve an
applied problem, sometimes only one answer satisfies the application.
Always check each answer in the words of the original problem.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 15
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
Step 1 Read the problem again. There will be two answers.
Step 2 Assign a variable. Let x represent the number of hours for the
slower carpet layer to complete the job alone. Then the faster carpet
layer could do the entire job in (x – 2) hours.
The slower person’s rate is , and the faster person’s rate is .1x
1x – 2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 16
Time Working Fractional Part
Rate Together of the Job Done
Slower Worker
Faster Worker
Step 2 (continued) Now complete the table below.
The slower person’s rate is , and the faster person’s rate is .
Together, they can do the job in 5 hr.
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
1x
1x – 2
1x
1x – 2
1x
1x – 2
(5)
(5)
5
5
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 17
Step 3 Write an equation. The sum of the fractional parts done by the
workers should equal 1 (the whole job).
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
5x
5x – 2
Part done by
slower worker
Part done by
faster worker+
+
=
=
1 whole job.
1
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 18
Step 4 Solve the equation from Step 3.
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
5x
5x – 2
+ = 1
Multiply by the LCD.5x
5x – 2
+ = 1x(x – 2) x(x – 2)
Distributive property+ =5(x – 2) x(x – 2)5x
Distributive property+ =5x – 10 x2 – 2x5x
Standard form= x2 – 12x + 100
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 19
Step 4 Solve the equation. (continued)
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
0 = x2 – 12x + 10
This equation cannot be solved by factoring, so use the quadratic
formula. (a = 1, b = –12, c = 10)
–b b2 – 4ac2a
x =+ 12 144 – 40
2x =
+
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 20
Step 4 Solve the equation. (continued)
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
or12 144 – 40
2x =
+
x ≈ 11.1
12 144 – 402
x =–
x ≈ .9 or
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 21
Step 5 State the answer. Only the solution 11.1 makes sense in the
original problem. (Why?) Thus, the slower worker can do the job in
about 11.1 hr and the faster in about 11.1 – 2 = 9.1 hr.
It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of
them could do the job in 2 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
EXAMPLE 3 Solving a Work Problem
10.3 Equations Quadratic in Form
Step 6 Check that these results satisfy the original problem.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 22
Solve each equation.
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
10.3 Equations Quadratic in Form
(a) n = –2n + 15
n2 = –2n + 15 Square both sides.
This equation is not quadratic. However, squaring both sides of the
equation gives a quadratic equation that can be solved by factoring.
n2 + 2n – 15 = 0 Standard form
(n + 5)(n – 3) = 0 Factor.
n + 5 = 0 or n – 3 = 0 Zero-factor property
n = –5 or n = 3 Potential solutions
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 23
Solve each equation.
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
10.3 Equations Quadratic in Form
(a) n = –2n + 15
Recall from Section 9.6 that squaring both sides of a radical equation can
introduce extraneous solutions that do not satisfy the original equation.
All potential solutions must be checked in the original (not the
squared) equation.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 24
Solve each equation.
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
10.3 Equations Quadratic in Form
(a) n = –2n + 15
Check: If n = –5, then
n = –2n + 15
–5 = –2(–5) + 15 ?
–5 = 25
If n = 3, then
n = –2n + 15
3 = –2(3) + 15 ?
3 = 9
–5 = 5 False 3 = 3 True
Only the solution 3 checks, so the solution set is { 3 }.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 25
Solve each equation.
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
10.3 Equations Quadratic in Form
(b) 3 e + e 10=
Isolate the radical on one side.3 e 10 – e=
Square both sides.9e 100 – 20e + e2=
Standard form0 e2 – 29e + 100=
Factor.0 (e – 4)(e – 25)=
Zero-factor propertye – 4 = 0 or e – 25 = 0
Potential solutionse = 4 or e = 25
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 26
Solve each equation.
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
10.3 Equations Quadratic in Form
(b) 3 e + e 10=
Check both potential solutions, 4 and 25, in the original equation.
Check: If e = 4, then If e = 25, then
3 e + e 10=
3 4 + 4 10 ?=
6 + 4 10 ?=
10 10 True=
3 e + e 10=
3 25 + 25 10 ?=
15 + 25 10 ?=
40 10 False=
Only the solution 4 checks, so the solution set is { 4 }.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 27
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(a) m4 – 26m2 + 25 = 0.
m4 – 26m2 + 25 = 0
Because m4 = (m2) 2, we can write this equation in quadratic form with
u = m2 and u2 = m4. (Instead of u, any letter other than m could be used.)
(m2)2 – 26m2 + 25 = 0 m4 = (m2)2
u2 – 26u + 25 = 0 Let u = m2.
(u – 1)(u – 25) = 0 Factor.
u = 1 or u = 25 Solve.
u – 1 = 0 or u – 25 = 0 Zero-factor property
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 28
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(a) m4 – 26m2 + 25 = 0.
To find m, we substitute m2 for u.
u = 1 or u = 25
m2 = 1 or m2 = 25
The equation m4 – 26m2 + 25 = 0, a fourth-degree equation, has
four solutions. * The solution set is { –5, –1, 1, 5 }. Check by substitution.
* In general, an equation in which an nth-degree polynomial equals 0 has
n solutions, although some of them may be repeated.
m = 1 or m = 5 Square root property+ +
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 29
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(b) c4 = 10c2 – 2.
c4 – 10c2 + 2 = 0 or ( c2 )2 – 10c2 + 2 = 0,
First write the equation as
u2 – 10u + 2 = 0.
which is quadratic in form with u = c2. Substitute u for c2 and u2 for c4 to get
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 30
Since this equation cannot be solved by factoring, use the quadratic formula.
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(b) c4 = 10c2 – 2.
u2 – 10u + 2 = 0
–b b2 – 4ac2a
x =+
10 100 – 82
u =+
a = 1, b = –10, c = 2
10 2 232
u =+
10 922
u =+
92 = 4 · =23 2 23
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 31
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(b) c4 = 10c2 – 2.
Factor.2 5 23
2u =
+
u = 5 23+ Lowest terms
c2 = 5 23+ or c2 = 5 23– Substitute c2 for u.
or Square root property5 23c = ++ 5 23c = –+
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 32
Solve each equation.
EXAMPLE 6 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
(b) c4 = 10c2 – 2.
The solution set contains four numbers:
5 23+ 5 23+– 5 23– 5 23––, , ,
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 33
10.3 Equations Quadratic in Form
Note on Solving Equations
NOTE
Some students prefer to solve equations like those in Example 6 (a) by
factoring directly. For example,
m4 – 26m2 + 25 = 0 Example 6(a) equation
(m2 – 1)(m2 – 25) = 0 Factor.
(m + 1)(m – 1)(m + 25)(m – 25) = 0. Factor again.
Using the zero-factor property gives the same solutions obtained in
Example 6(a). Equations that cannot be solved by factoring (as in
Example 6(c)) must be solved by substitution and the quadratic formula.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 34
Solve 3(2x – 1)2 + 8(2x – 1) – 35 = 0.
EXAMPLE 7 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
3(2x – 1)2 + 8(2x – 1) – 35 = 0
Because of the repeated quantity 2x – 1, this equation is quadratic in
form with u = 2x – 1.
3u2 + 8u – 35 = 0 Let 2x – 1 = u.
(3u – 7)(u + 5) = 0 Factor.
3u – 7 = 0 or u + 5 = 0 Zero-factor property
u = or u = –5 Zero-factor property73
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 35
Solve 3(2x – 1)2 + 8(2x – 1) – 35 = 0.
EXAMPLE 7 Solving Equations That are Quadratic
in form
10.3 Equations Quadratic in Form
u = or u = –573
2x – 1 = or 2x – 1 = –5 Substitute 2x – 1 for u.73
2x = or 2x = –4 Solve for x. 103
x = or x = –2 Solve for x. 53
Check that the solution set of the original equation is –2, .53
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 36
10.3 Equations Quadratic in Form
Caution
CAUTION
A common error when solving problems like those in Examples 6 and 7 is to
stop too soon. Once you have solved for u, remember to substitute and
solve for the values of the original variable.