Equations Quadratic in Form

Contributed by:

Sharp Tutor Mon, Jan 17, 2022 07:29 PM UTC

OBJECTIVE:
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.

1. Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 10.3 - 1

2. Chapter 10
Quadratic Equations,
Inequalities,
and Functions
Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 10.3 - 2

3. 10.3
Equations Quadratic in Form
Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 10.3 - 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 - 4

5. 10.3 Equations Quadratic in Form
EXAMPLE 1 Solving an Equation with Fractions That
Leads to a Quadratic Equation
Solve 3 2 5 .
+ =
x x–5 6
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.)
3 2 5
6x(x – 5) + 6x(x – 5) = 6x(x – 5)
x x–5 6
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 - 5

6. 10.3 Equations Quadratic in Form
EXAMPLE 1 Solving an Equation with Fractions That
Leads to a Quadratic Equation
Solve 3 2 5 .
+ =
x x–5 6
Combine and rearrange terms so that the quadratic equation is in standard
form. Then factor to solve the resulting equation.
30x – 90 = 5x2 – 25x
5x2 – 55x + 90 = 0 Standard form
5(x2 – 11x + 18) = 0 Factor.
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 - 6

7. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
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 sp
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 - 7

8. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
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 sp
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 - 8

9. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
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 sp
of the current, or 10 – x.
d r t
Upstream 8 10 – x 8
10 – x
8
Downstream 8 10 + x 10 + x
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 9

10. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
Step 3 Write an equation. The total time, 1 hr and 40 min, can be written as
1 + 40 =+ 1 + 2 = = 5
60 3 3 hr.
Time upstream Time downstream Total Time
d r t
Upstream 8 10 – x 8
10 – x
8
Downstream 8 10 + x 10 + x
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 10

11. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
Step 4 Solve the equation. Multiply each side by 3(10 – x)(10 + x), the LCD,
and solve the resulting quadratic equation.
8 + 8 = 5
10 – x 10 + x 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 - 11

12. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
Step 4 Solve the equation. Multiply each side by 3(10 – x)(10 + x), the LCD,
and solve resulting quadratic equation.
240 + 24x + 240 – 24x = 500 – 5x2
480 = 500 – 5x2 Combine terms.
5x2 = 20
x2 = 4 Divide by 5.
x = 2 or x = –2 Square root property
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 12

13. 10.3 Equations Quadratic in Form
EXAMPLE 2 Solving a Motion Problem
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.
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 8
–2x 1
Downstream 8 1012
+2x 2
3
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 13

14. 10.3 Equations Quadratic in Form
Caution on “Solutions”
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 - 14

15. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
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 car
layer could do the entire job in (x – 2) hours.
1 is
The slower person’s rate , and the faster person’s rate1is
x x–2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 15

16. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
Step 2 (continued) Now complete the table below.
1 is
The slower person’s rate , and the faster person’s rate1is
x x–2
Together, they can do the job in 5 hr.
Time Working Fractional Par
Rate Together of the Job Don
1 1 (5)
Slower Worker x 5 x
1 5 1 (5)
Faster Worker x–2 x–2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 16

17. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
Step 3 Write an equation. The sum of the fractional parts done by the
workers should equal 1 (the whole job).
Part done by Part done by
+ = 1 whole job.
slower worker faster worker
5 + 5 = 1
x x–2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 17

18. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
Step 4 Solve the equation from Step 3.
5 + 5 = 1
x x–2
x(x – 2) 5 + 5 = x(x – 2) 1 Multiply by the LCD.
x x–2
5(x – 2) + 5x = x(x – 2) Distributive property
5x – 10 + 5x = x2 – 2x Distributive property
0 = x2 – 12x + 10 Standard form
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 18

19. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
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)
x = –b + b2 – 4ac x =
12 + 144 – 40
2a 2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 19

20. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
Step 4 Solve the equation. (continued)
12 + 144 – 40 12 – 144 – 40
x = or x =
2 2
x ≈ 11.1 or x ≈ .9
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 20

21. 10.3 Equations Quadratic in Form
EXAMPLE 3 Solving a Work 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?
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 i
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.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 21

22. 10.3 Equations Quadratic in Form
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
Solve each equation.
(a) n = –2n + 15
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 Square both side
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 solution
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 22

23. 10.3 Equations Quadratic in Form
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
Solve each equation.
(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 - 23

24. 10.3 Equations Quadratic in Form
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
Solve each equation.
(a) n = –2n + 15
Check: If n = –5, then If n = 3, then
n = –2n + 15 n = –2n + 15
–5 = –2(–5) + 15 ? 3 = –2(3) + 15 ?
–5 = 25 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 - 24

25. 10.3 Equations Quadratic in Form
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
Solve each equation.
(b) 3 e + e = 10
3 e = 10 – e Isolate the radical on one side.
9e = 100 – 20e + e2 Square both sides.
0 = e2 – 29e + 100 Standard form
0 = (e – 4)(e – 25) Factor.
e – 4 = 0 or e – 25 = 0 Zero-factor property
e = 4 or e = 25 Potential solutions
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 25

26. 10.3 Equations Quadratic in Form
EXAMPLE 4 Solving Radical Equations That Lead to
Quadratic Equations
Solve each equation.
(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 e + e = 10
3 4 + 4 = 10 ? 3 25 + 25 = 10 ?
6 + 4 = 10 ? 15 + 25 = 10 ?
10 = 10 True 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 - 26

27. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(a) 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.)
m4 – 26m2 + 25 = 0
(m2)2 – 26m2 + 25 = 0 m4 = (m2)2
u2 – 26u + 25 = 0 Let u = m2.
(u – 1)(u – 25) = 0 Factor.
u – 1 = 0 or u – 25 = 0 Zero-factor property
u = 1 or u = 25 Solve.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 27

28. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(a) m4 – 26m2 + 25 = 0.
To find m, we substitute m2 for u.
u = 1 or u = 25
m2 = 1 or m2 = 25
m = +1 or m = +5 Square root property
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.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 28

29. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(b) c4 = 10c2 – 2.
First write the equation as
c4 – 10c2 + 2 = 0 or ( c2 )2 – 10c2 + 2 = 0,
which is quadratic in form with u = c2. Substitute u for c2 and u2 for c4 to get
u2 – 10u + 2 = 0.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 29

30. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(b) 4 2 x = –b + b2 – 4ac
c = 10c – 2. 2a
Since this equation cannot be solved by factoring, use the quadratic formula.
u2 – 10u + 2 = 0
10 + 100 – 8
u = a = 1, b = –10, c = 2
2
10 + 92
u =
2
10 + 2 23
u = 92 = 4 · 23 = 2 23
2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 30

31. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(b) c4 = 10c2 – 2.
2 5 + 23
u = Factor.
2
u = 5 + 23 Lowest terms
c2 = 5 + 23 or c2 = 5 – 23 Substitute c2 for u.
c = + 5 + 23 or c = + 5 – 23 Square root property
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 31

32. 10.3 Equations Quadratic in Form
EXAMPLE 6 Solving Equations That are Quadratic
in form
Solve each equation.
(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 - 32

33. 10.3 Equations Quadratic in Form
Note on Solving Equations
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 - 33

34. 10.3 Equations Quadratic in Form
EXAMPLE 7 Solving Equations That are Quadratic
in form
Solve 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.
3(2x – 1)2 + 8(2x – 1) – 35 = 0
3u2 + 8u – 35 = 0 Let 2x – 1 = u.
(3u – 7)(u + 5) = 0 Factor.
3u – 7 = 0 or u + 5 = 0 Zero-factor property
7
u = or u = –5 Zero-factor property
3
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 34

35. 10.3 Equations Quadratic in Form
EXAMPLE 7 Solving Equations That are Quadratic
in form
Solve 3(2x – 1)2 + 8(2x – 1) – 35 = 0.
7
u = 3 or u = –5
7
2x – 1 = 3 or 2x – 1 = –5 Substitute 2x – 1 for u.
10
2x = or 2x = –4 Solve for x.
3
5
x = or x = –2 Solve for x.
3
5
Check that the solution set of the original equation is –2, .
3
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 35

36. 10.3 Equations Quadratic in Form
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 - 36

Book a Trial Class

Download