Precalculus
Notes: Solving Rational Equations, Radical Equations
Part I: Solving Rational Equations
State the domain, solve for the variable and circle final answers.
Name: _______________________________________
1/8/14
Domain:____________ 2. x
2 x

3
 x
4


3 9
24
 x
2
Domain: ____________ 1. x x

2
 x
1

2
 x
2
8

4
3. x x

3
 x
4

3
 x
2
18

9
Domain:_______________ 4. x x

2
 x x


2 4

8 x 2
Domain: ____________
Part II – Equations Involving Radicals
**Always isolate the radical before solving for the variable
State the domain, solve for x and check your answers for extraneous solutions.
1. 2 x 7 x 2 Domain: x


2
7
2. 2 x 6 x 4 1 Domain: x
 
3
Check :
2 x 6 x 4 1
2 x
 
 x

2
2 x 7 2
2 x
 
( x

2)
2
2 x 7 x
2

4 x

4 x
2

2 x
 
0
( x

3)( x
 
0 x
 
3 x

1
2( 3) 7 ( 3) 2
1 3 2
Check :
2(1) 7 1 2
2 x 6 ( x

4)

2 x 1 2 x

4
( x

1)
2 
4( x

4) x
2

2 x 1 4 x

16 x
2

2 x

15

0
( x

5)( x
 
0 x

5 x
 
3 x
Check by graph

Part III – Equations Involving Absolute Value
**Always isolate the absolute value before solving for the variable
1. x
2

3 x

4 x

6 Check 2. x

10
 x
2

10 x

0 x
2 
3 x
 
4 x x
 
3 and 1 x

10
 x 2

10 x
Check x
 
1 and 10
First Equation : x 2

3 x
 
4 x x 2 x 6 0
( x

3)( x

2)

0 x
 
3 x

2
First Equation : x

10
 x
2

10 x x 2

11 x

10

0
( x

10)( x
 
0 x

10 x

1
Second Equation x 2

3 x (6 4 ) x
2 
3 x

4 x

6 x
2 
7 x
 
0
( x

6)( x
 
0 x

6 x

1
Second Equation : x

10
 
( x 2
 x x

10 x 10 x x
2

9 x

10

0
( x

10)( x
 
0 x

10 x
 
1
Part IV – Absolute Value Inequalities
A. Equation/Inequality x

2 x x


2
2 x

2
Solution
OR x 2
Verbal Description
 
Values of x that lie 2 units from
0. x

2 x 2
OR x
AND Values of x that lie less than 2
2 units from 0.
 
Values of x that lie more than
2 units from 0.
B. Examples: Solve and graph the solution
1. 2 x 1 7 Solution set: ( 4, 3)
( , 4) (3, )
 
2 x
 
 
2 x

6 x 3
2. 2 x 1 7
2 x
 
7
2 x

6 x

3
Graph
Solution Set:
OR 2 x
  
7
2 x x
 
8
 
4
3. 4 2 x 5 36 Solution set: ( , 2] [7, ) 4.
1
2
2 x
 
9
 x
 
9 Solution set:


8,
40
3


 x

32
2 x
 
9
2 x

14 x

7
OR 2 x
  
9
2 x
 
4 x
 
2
   x

32
   
24
40
3 x 8 8 x
40
3

Part V – Polynomial Inequalities
A. Between two consecutive zeros a polynomial must be entirely positive or entirely negative (above the x-axis
or below the x-axis). This means that when the real zeros of a polynomial are put in order, they divide the
real line into intervals in which the polynomial has no sign changes.
B. These zeros are called critical values. The intervals are called test intervals.
C. Example: x 2 
2 x
 
0  x
2 
2 x
 
( x

3)( x

1) which means that the zeros are x = -1 and x = 3.
The zeros divide the number line into three test intervals: (− ∞ , −1), (−1, 3), 𝑎𝑛𝑑 (3, ∞ )
D. **To solve an inequality, you only need to test one value from each of the test intervals.
E. Test values for above example: x = -2, x = 0, x = 4
 2    
0 ?
(0)
2   
0 ?
(4)
2   
0 ?
4 4 3 0 ?
5

0
***The solution set is (-1, 3).
0 0 3
3 0
0 ?
16 8 3
5

0
0 ?
F. Examples: Find the critical values and use the test intervals to determine the solution.
1. 2 x x 6 0 2. 2 x
2 
5 x

12 3. x
2 
3 x
 
5
Critical values: x = -2, x = 3 Critical values: x = -4, x =
3
2
Critical values: No real solutions
Part VI – Rational Inequalities
A. Similar to polynomial inequalities, the value of a rational expression can change signs only at is zeros and its
undefined values. These two types of numbers make up the critical values.
B. Example: Find the critical values and the test intervals to determine the solution.
1. x x


12
2

3 x x


12
2
3 0 Critical Values:
  
0 x

3 x
  x

6 x


2 x

6
2
 x

2
0

0 Solution set: (-2, 3] x x
2 0
 
2
2.
2 x

7 x

5

3
2 x

7 x

5
3 0
2 x 7 3 x

15 x

5 x

5
8

0

0
3. x
1

3

4 x
9

3 x
(
1
 x
3

4 x
9

3

3)(4 x

3)

4 x 3 9 x

27
0

 
30
( x

3)(4 x

3)

0
0
Critical Values: 𝒙 = 𝟓 𝒂𝒏𝒅 𝒙 = 𝟖
Solution Set: (− ∞ , 5) ∪ (8, ∞ )
Critical Values: 𝒙 = 𝟔, 𝒙 = 𝟑 𝒂𝒏𝒅 = 𝒙 −
𝟑
𝟒
Solution Set: (−
3
4
, 3) ∪ (6, ∞ )
Homework: Pg 143 #33, 38, 41, 46
Pg 179 #57, 62, 64, 67, 74, 86
Pg 192 # 20, 30, 55, 61, 64, 69, 72