* Solve equations involving rational exponents * Solve equations using factoring * Solve equations with radicals and check the solutions * Solve absolute value equations * Solve other types of equations

1. 2.6 Other Types of Equations
Chapter 2 Equations and Inequalities

2. Concepts and Objectives
⚫ Objectives for this section are:
⚫ Solve equations involving rational exponents
⚫ Solve equations using factoring
⚫ Solve equations with radicals and check the solutions
⚫ Solve absolute value equations
⚫ Solve other types of equations

3. Rational Exponent Equations
⚫ Recall that a rational exponent indicates a power in the
numerator and a root in the denominator.
⚫ There are multiple ways of writing an expression with a
rational exponent:
⚫ If we are given an equation in which a variable is raised
to a rational exponent, the simplest way to remove the
exponent on x is by raising both sides of the equation to
a power that is the reciprocal of the exponent.
( ) ( )
/ 1/
m
m
m n n m
n n
a a a a
= = =

4. Rational Exponent Equations
⚫ Example: Solve
The reciprocal of is , so
5/4
32
x =
5
4
4
5
( ) ( )
4/5 4/5
5/4
32
16
x
x
=
=

5. Rational Exponent Equations
⚫ Example: Solve
We can now use the zero-factor property:
3/4 1/2
3x x
=
( )
3/4 1/2
3/4 2/4
2/4 1/4
3 0
3 0
3 1 0
x x
x x
x x
− =
− =
− =

6. Rational Exponent Equations
⚫ Example: Solve
Solutions are 0 and
3/4 1/2
3x x
=
2/4
0
0
x
x
=
=
1/4
1/4
1/4
4
3 1 0
3 1
1
3
1 1
3 81
x
x
x
x
− =
=
=
 
= =
 
 
1
81

7. Solving by Grouping
⚫ If a polynomial consists of 4 terms, we can solve it by
grouping.
⚫ Grouping procedures require factoring the first two
terms, and then factoring the last two terms. If the
factors in the parentheses are identical, then the
expression can be factored by grouping.

8. Solving by Grouping
⚫ Example: Solve
Solutions are ‒1, 3, ‒3
3 2
9 9 0
x x x
+ − − =
( ) ( )
( )( )
( )( )( )
3 2
2
2
9 9 0
1 9 1 0
1 9 0
1 3 3 0
x x x
x x x
x x
x x x
+ − − =
+ − + =
+ − =
+ − + =
difference of squares!

9. Power Property
⚫ Note: This does not mean that every solution of Pn = Qn
is a solution of P = Q.
⚫ We use the power property to transform an equation
that is difficult to solve into one that can be solved more
easily. Whenever we change an equation, however, it is
essential to check all possible solutions in the original
equation.
If P and Q are algebraic expressions, then every
solution of the equation P = Q is also a solution of
the equation Pn = Qn, for any positive integer n.

10. Solving Radical Equations
⚫ Step 1 Isolate the radical on one side of the equation.
⚫ Step 2 Raise each side of the equation to a power that is
the same as the index of the radical to eliminate the
radical.
⚫ If the equation still contains a radical, repeat steps 1
and 2.
⚫ Step 3 Solve the resulting equation.
⚫ Step 4 Check each proposed solution in the original
equation.

11. Solving Radical Equations (cont.)
⚫ Example: Solve − + =
4 12 0
x x

12. Solving Radical Equations (cont.)
⚫ Example: Solve − + =
4 12 0
x x
= +
4 12
x x
= +
2
4 12
x x
− − =
2
4 12 0
x x
( )( )
− + =
6 2 0
x x
= −
6, 2
x

13. Solving Radical Equations (cont.)
⚫ Example: Solve
Check:
Solution: {6}
− + =
4 12 0
x x
4 12
x x
= +
2
4 12
x x
= +
2
4 12 0
x x
− − =
( )( )
6 2 0
x x
− + =
6, 2
x = −
( )
− + =
6 4 6 12 0
− =
6 36 0
− =
6 6 0
=
0 0
( )
− − − + =
2 4 2 12 0
− − =
2 4 0
− − =
2 2 0
− 
4 0

14. Solving Radical Equations (cont.)
⚫ Example: Solve + − + =
3 1 4 1
x x

15. Solving Radical Equations (cont.)
⚫ Example: Solve + − + =
3 1 4 1
x x
( )
2
4 1
x + +
( )
2
2
3 1 4 1
3 1 4 2 4 1
2 4 2 4
2 4
4 4 4
5 0
5 0
0, 5
x x
x x x
x x
x x
x x x
x x
x x
x
+ = + +
+ = + + + +
− = +
− = +
− + = +
− =
− =
=

16. Solving Radical Equations (cont.)
⚫ Example: Solve
Check:
Solution: {5}
+ − + =
3 1 4 1
x x
( )
2
2
3 1 4 1
3 1 4 2 4 1
2 4 2 4
2 4
4 4 4
5 0
5 0
0, 5
x x
x x x
x x
x x
x x x
x x
x x
x
+ = + +
+ = + + + +
− = +
− = +
− + = +
− =
− =
=
( )+ − + =
3 0 1 0 4 1
− =
1 4 1
− =
1 2 1
− 
1 1
( )+ − + =
3 5 1 5 4 1
− =
16 9 1
− =
4 3 1
=
1 1

17. Absolute Value Equations
⚫ You should recall that the absolute value of a number a,
written |a|, gives the distance from a to 0 on a number
line.
⚫ By this definition, the equation |x| = 3 can be solved by
finding all real numbers at a distance of 3 units from 0.
Both of the numbers 3 and ‒3 satisfy this equation, so
the solution set is {‒3, 3}.

18. Absolute Value Equations (cont.)
⚫ The solution set for the equation must include
both a and –a.
⚫ Example: Solve
=
x a
− =
9 4 7
x

19. Absolute Value Equations (cont.)
⚫ The solution set for the equation must include
both a and –a.
⚫ Example: Solve
The solution set is
=
x a
− =
9 4 7
x
− =
9 4 7
x − = −
9 4 7
x
− = −
4 2
x − = −
4 16
x
=
1
2
x = 4
x
or
 
 
 
1
,4
2

20. Quadratic in Form
⚫ An equation is said to be quadratic in form if it can be
written as
where a  0 and u is some algebraic expression.
⚫ To solve this type of equation, substitute u for the
algebraic expression, solve the quadratic expression for
u, and then set it equal to the algebraic expression and
solve for x. Because we are transforming the equation,
you will still need to check any proposed solutions against
the original equation.
+ + =
2
0
au bu c

21. Quadratic in Form (cont.)
⚫ Example: Solve ( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x

22. Quadratic in Form (cont.)
⚫ Example: Solve
Let This makes our equation:
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( )
= −
1 3
1
u x
+ − =
2
12 0
u u
( )( )
+ − =
4 3 0
u u
= −4, 3
u

23. Quadratic in Form (cont.)
⚫ Example: Solve
Let . This makes our equation:
So, and
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( )
= −
1 3
1
u x
+ − =
2
12 0
u u
( )( )
+ − =
4 3 0
u u
= −4, 3
u
( )
( ) ( )
1 3
3
1/3 3
1 4
1 4
1 64
63
x
x
x
x
− = −
 
− = −
 
− = −
= −
( )
( ) ( )
1 3
3
1/3 3
1 3
1 3
1 27
28
x
x
x
x
− =
 
− =
 
− =
=

24. Quadratic in Form (cont.)
⚫ Example: Solve (cont.)
Now, we have to check our proposed solutions:
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( ) ( )
− − + − − − =
2 3 1 3
63 1 63 1 12 0
( ) ( )
− + − − =
2 3 1 3
64 64 12 0
( ) ( )
− + − − =
2 1
4 4 12 0
− − =
16 4 12 0
=
0 0

25. Quadratic in Form (cont.)
⚫ Example: Solve (cont.)
Solution: {–63, 28}
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( ) ( )
− + − − =
2 3 1 3
28 1 28 1 12 0
( ) ( )
+ − =
2 3 1 3
27 27 12 0
( ) ( )
+ − =
2 1
3 3 12 0
+ − =
9 3 12 0
=
0 0

26. Rational Equations (Again)
⚫ A rational equation is an equation that has a rational
expression for one or more terms.
⚫ To solve a rational equation, multiply both sides by the
lowest common denominator of the terms of the
equation. Be sure to check your solution against the
undefined values!
Because a rational expression is not defined when its
denominator is 0, any value of the variable which makes
the denominator’s value 0 cannot be a solution.

27. Rational Equations (cont.)
⚫ Example: Solve 2
3 2 1 2
2 2
x
x x x x
+ −
+ =
− −

28. Rational Equations (cont.)
⚫ Example: Solve
The LCD is xx ‒ 2, which means x  0, 2.
2
3 2 1 2
2 2
x
x x x x
+ −
+ =
− −
( ) ( ) ( )
( )
2
2 2
3 2 1 2
2
2
x
x x x x
x
x x x x x
 
+ −
   
+ =  
     
−
− − −
−
     

29. Rational Equations (cont.)
⚫ Example: Solve
The LCD is xx ‒ 2, which means x  0, 2.
2
3 2 1 2
2 2
x
x x x x
+ −
+ =
− −
( ) ( ) ( )
( )
( ) ( )
( )
2
2
3 2 1 2
2 2
3 2 2 2
3 2 2 2
2
3 3 0
3 1 0
2 2
0, 1
x
x
x x x x
x x x
x
x
x x x x
x x
x x
x x
x
 
+ −
   
+ =  
     
− −
     
+ + − = −
+ + − = −
+ =
+ =
=
− −
−
−
 
1
−

30. Classwork
⚫ College Algebra 2e
⚫ 2.6: 6-18 (even); 2.5: 20-30 (even); 2.4: 28-40 (even)
⚫ 2.6 Classwork Check
⚫ Quiz 2.5