copyright © 2010 pearson education, inc. all rights reserved sec 10.3 - 1


Quadratic Equations, Inequalities,

and Functions

Chapter 10


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.


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


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.


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


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 }.


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


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.










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.










810 – x

810 + x

d r t

Upstream 8 10 – x

Downstream 8 10 + x


=

810 – x

810 + x





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






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






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






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



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.


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


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


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.






1x

1x – 2

1x

1x – 2

1x

1x – 2

(5)

(5)

5

5


Step 3 Write an equation. The sum of the fractional parts done by the

workers should equal 1 (the whole job).






5x

5x – 2

Part done by

slower worker

Part done by

faster worker+

+

=

=

1 whole job.

1


Step 4 Solve the equation from Step 3.






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


Step 4 Solve the equation. (continued)






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 =

+


Step 4 Solve the equation. (continued)






or12 144 – 40

2x =

+

x ≈ 11.1

12 144 – 402

x =–

x ≈ .9 or


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.






Step 6 Check that these results satisfy the original problem.


Solve each equation.

EXAMPLE 4 Solving Radical Equations That Lead to

Quadratic Equations


(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




Quadratic Equations


(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.




Quadratic Equations


(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 }.




Quadratic Equations


(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




Quadratic Equations


(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 }.



EXAMPLE 6 Solving Equations That are 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




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+ +




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


Since this equation cannot be solved by factoring, use the quadratic formula.



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




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 = –+




in form


(b) c4 = 10c2 – 2.

The solution set contains four numbers:

5 23+ 5 23+– 5 23– 5 23––, , ,



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.


Solve 3(2x – 1)2 + 8(2x – 1) – 35 = 0.


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


Solve 3(2x – 1)2 + 8(2x – 1) – 35 = 0.


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



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.

copyright © 2010 pearson education, inc. all rights reserved sec 10.3 - 1

Documents

x slide

x downstream8

x drt upstream8

quadratic form

quadratic equations

equations quadratic

form slide

pearson education