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Simplifying Radical Expressions
Multiply.
23 = 6
2 3 = 2 3
2 3 6
Multiply.
5  7 =35
5 7 = 5 7
5  7  35
Simplify the following:
20, 40, 50, 48, 120, 85, 200, 150,
3
16, 3 54, 3 128, 3 256, 4 32, 4 64, 4 162, 4 512,
5
64, and 5 128.
Solution :
20  4  5  4  5  2 5
40  4  10  4  10  2 10
50  25  2  25  2  5 2
48  16  3  16  3  4 3
120  4  30  4  30  2 30
85  85
200  100  2  100  2  10 2
150  25  6  25  6  5 6
3
16  3 8  2  3 8  3 2  2 3 2
3
54  3 27  2  3 27  3 2  3 3 2
3
128  3 64  2  3 64  3 2  4 3 2
3
256  3 64  4  3 64  3 4  4 3 4
4
32  4 16  2  4 16  4 2  2 4 2
4
64  4 16  4  4 16  4 4  2 4 4
4
162  4 81  2  4 81  4 2  3 4 2
4
512  4 256  2  4 256  4 2  4 4 2
5
64  5 32  2  5 32  5 2  2 5 2
5
128  5 32  4  5 32  5 4  2 5 4
Simplify the following:
20 x 2
Solution :
20 x 2  20 x 2  4  5 x 2  4 5 x 2  2  5  x  2 x 5
Simplify the following:
45 x 5
Solution :
45 x 5  9  5 x 2  x 2  x1  9 5 x 2 x 2 x
 3  5  x  x  x  3x 2 5 x
 3x2 5x
Simplify the following:
80 x 5 y 6
Solution :
80 x 5 y 6  16  5 x 2  x 2  x1  y 2  y 2  y 2
 16 5 x 2 x 2 x y 2 y 2 y 2
= 4 5  x x x  y y y
 4 x2 y3 5 x
 4 x2 y3 5x
Simplify the following:
3
80 x 5 y 6
Solution :
3
80 x 5 y 6  3 8  10 x 3 x 2 y 3 y 3
= 3 8  3 10  3 x 3  3 x 2  3 y 3  3 y 3
= 2  3 10  x  3 x 2  y  y
= 2y 2 x  3 10  3 x 2
= 2xy 2 3 10 x 2
Simplify the following:
8 5
a
27
Solution :
8 5
8  a2  a2  a
a 

27
27
=
=
4  2  a2  a2  a
93
4 2 a2 a2 a
9 3
2 2aa a
3 3
2  a 2  2a
=
3 3
Simplify the following:
3
8 10 10
a b
27
Solution :
3
8 10 10 3 8  a 3  a 3  a 3  a  b3  b3  b3  b
a b 
3
27
27
3
8  3 a 3  3 a 3  3 a 3  3 a  3 b3  3 b3  3 b3  3 b
3
27
3
8  3 a 3  3 a 3  3 a 3  3 a  3 b3  3 b3  3 b3  3 b
3
27
=
=
2a a a  3 a bbb 3 b
=
3
2  a 3  b3  3 a  3 b
=
3
2  a 3  b3  3 ab
=
3
Simplify the following:
5
a 22b15
Solution :
5
a 22b15  5 a 5  a 5  a 5  a 5  a 2  b5  b5  b5
= 5 a 5  5 a 5  5 a 5  5 a 5  5 a 2  5 b5  5 b5  5 b5
= a  a  a  a  5 a2  b  b  b
= a 4  b3  5 a 2
Right Triangle Problems
a
c
b
a2 + b2 = c 2
(leg 1)2 + (leg 2)2 =  hypotenuse 
2
Find the missing side.
x
4
8
Solution :
4 2  82  x 2
16  64  x 2
80  x 2
x 2  80
x 2  80
x  16  5
x  16 5  4 5
Find the missing side.
10
4
b
Solution :
42  b 2  102
16  b 2  100
16  b 2  16  100  16
b 2  84
b 2  84
b  4  21
b  4 21  2 21
Distance Between Two Points
Distance Between Two Points =
 x2  x1    y2  y1 
2
2
 x1 , y1 
 x2 , y2 
Example 1
Find the distance between (4, 5) and (11, 3)
Solution :
x1  4; y1  5; x2  11; y2  3
Distance Between Two Points =
 x2  x1 
Distance Between Two Points =
11  4 
Distance Between Two Points =
7 
2
2
2
  y2  y1 
 3  5
  2 
2
2
2
Distance Between Two Points = 49  4
Distance Between Two Points = 53  7.28
Midpoint of the segment joining (4, 5) and (11, 3):
 x  x y  y2   4  11 5  3 
Midpoint =  1 2 , 1
,
   7.5, 4 

2   2
2 
 2
Example 2
Find the distance between (-4, 5) and (-1, 8)
Solution :
x1  4; y1  5; x2  1; y2  8
Distance Between Two Points =
 x2  x1 
Distance Between Two Points =
 (1)  (4) 
Distance Between Two Points =
 3
2
2
  3
  y2  y1 
2
2
 8  5 
2
2
Distance Between Two Points = 9  9
Distance Between Two Points = 18= 9  2=3 2  4.24
Midpoint of the segment joining (-4, 5) and (-1, 8):
 x  x y  y2   4  (1) 5  8 
Midpoint =  1 2 , 1
,
   2.5, 6.5 

2
2
2
2


 
Example 3
Find the distance between (4.2, -5.7) and (-1.8, 8.2)
Solution :
x1  4.2; y1  5.7; x2  1.8; y2  8.2
Distance Between Two Points =
 x2  x1 
Distance Between Two Points =
 (1.8)  (4.2) 
Distance Between Two Points =
 6 
2
2
  y2  y1 
 13.9 
2
2
  8.2  (5.7) 
2
2
Distance Between Two Points = 229.21  15.14
Midpoint of the segment joining (4.2, -5.7) and (-1.8, 8.2):
 x  x y  y2   4.2  (1.8) (5.7)  8.2 
Midpoint =  1 2 , 1
,
  1.2, 1.25 

2
2
2
2


 
Example 4
Find the distance between ( 5, 10) and ( 14, 19)
Solution :
x1  5; y1  10; x2  14; y2  19
Distance Between Two Points =
 x2  x1 
Distance Between Two Points =

Distance Between Two Points =
1.5055894 
2
  y2  y1 
14  5
 
2

2
2
19  10

2
 1.196621283
Distance Between Two Points = 3.698701965  1.923
2