Download 1.2 Exponents and Powers

Factor.
2
2x –3
3x  5
–10
––55 +22
(2x  5)(
)

2x  1
Check using FOIL.
6x2  9x  15
3 2x 2  3x  5


-10
- -5
5
+2
-3
(2x  5 )(
)
2 x  1
Check using FOIL.
If you factor out a GMF:
before using the X figure, keep it,
after using the X figure, discard it.
10-3A Radical Equations
Algebra 1
Glencoe McGraw-Hill
Linda Stamper
What inverse operation is needed to solve
each of the following?
x 3  7
addition
x  3  10
subtraction
2x  12
x
3
4
x2  4
division
x 3
squaring
multiplication
square root
When solving radical equations you must square both
sides of the equation to remove the square root sign
(undo the radical symbol).
x 8
 x 2  82
x  64
Squaring is the
inverse
operation of
square root!
When solving radical equations you must square both sides
of the equation to remove the square root sign (undo the
radical symbol).
Check
x 8
x 8
 x 2  82
x  64
64  8
8 8
CAUTION: Squaring both sides of an equation can
introduce a solution to the squared equation that does not
satisfy the original equation. Such a solution is called an
extraneous solution. When you solve by squaring both
sides of an equation, check each solution in the original
equation.
Solve.
Example 1
Example 2
Example 3
x 1  5
x  10  0
x 5  0
Example 4
8 x  3  16
Isolate the radical
on one side of the
equal sign before
you undo the
radical!
Example 5
3x  1  2  6
Solve.

Example 1
Example 2
Example 3
x 1  5
x  10  0
 10  10
x  10
x 5  0
5 5
x  1   5 2
2
x  1  25
1 1
x  24

2
x   10 2
x  100
x  5
 x 2   52
x  25
No solution
Solve.
Example 5
Example 4
8 x  3  16
8
8
x 3  2

x  3   22
2
x 3  4
3 3
x7
3x  1  2  6
2 2
3x  1  8

3x  1   82
3x  1  64
1 1
2
3x  63
3
3
x  21
Okay to keep
fractional answers
improper, but in
lowest terms!
Solve.
1)
2x  5  13

2x  5   132
2
2x  5  169
5
5
2x  174
2
2
x  87
2)
4  2x  3  5
4
4
 2x  3  1
1
1
2x  3  1
No Solution
Solve.
3)
x  5  10  8
 10  10
4)
5  4x  3  3
5
5
x 5  2

 4x  3  2
1
1
x  5   22
2
x 5  4
5 5
x9
4x  3  2

4x  3   22
2
4x  3  4
3 3
4x  7
7
x
4
Solve.
5)
4x
9 3
5
9 9
4x
 12
5
2
 4x 

  12 2
 5 
 5  4x
5
 144  
4
  5
4
x  180
6)

3x  2  2 3
3x  2   2 3 
3x  2  43
2
3x  2  12
3x  10
10
x
3
2
Solve.
7)
8)
2
x
x  8x  31  4
2

2
2
 8x  31  4 
x2  8x  31  16
x2  8x  15  0
x  5x  3  0
15
5 3
8
x 5  0 x 3  0
x  3
x  5
2x2  21x  7  2
 2x

2
 21x  7  22
2x2  21x  7  4
2x2  21x  11  0
2
-22
+ 22
--1
1
21
(
)( 2x  1)  0
2x  11
0
x  11  0 2x  1  0
1
 11,
2
Solve.
8
9))
2
4 2
x 3  
5
5 5
4 4
 
5 5
2
6
x 3 
5
5
2
2
 2

6


x  3    
5
 5

2
36
x 3 
5
25
75

 3 25 1
 39  5 
5 2
x

 
 
25  2 
2 5
5
x
39
10
10-A10 Pages 544-546 #11-22,70-75.