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To simplify a square root expression that does not involve a fraction, we can use the following two rules:

Simplifying square roots without fractions

  1. If a factor of the radicand contains a variable with an even exponent, the square root is obtained by dividing the exponent by 2.
  2. If a factor of the radicand contains a variable with an odd exponent, the square root is obtained by first factoring the variable factor into two factors so that one has an even exponent and the other has an exponent of 1, then using the product property of square roots.

Sample set a

Simplify each square root.

a 4 .     The exponent is even: 4 2 = 2. The exponent on the square root is 2.

a 4 = a 2

a 6 b 10 .    Both exponents are even: 6 2 = 3 and 10 2 = 5. The exponent on the square root of a 6 is 3. The exponent on the square root if b 10 is 5.

a 6 b 10 = a 3 b 5

y 5 .    The exponent is odd: y 5 = y 4 y . Then

y 5 = y 4 y = y 4 y = y 2 y

36 a 7 b 11 c 20 = 6 2 a 6 a b 10 b c 20 a 7 = a 6 a , b 11 = b 10 b = 6 2 a 6 b 10 c 20 · a b by the commutative property of multiplication. = 6 2 a 6 b 10 c 20 a b by the product property of square roots. = 6 a 3 b 5 c 10 a b

49 x 8 y 3 ( a 1 ) 6 = 7 2 x 8 y 2 y ( a 1 ) 6 = 7 2 x 8 y 2 ( a 1 ) 6 y = 7 x 4 y ( a 1 ) 3 y

75 = 25 · 3 = 5 2 · 3 = 5 2 3 = 5 3

Practice set a

Simplify each square root.

m 8

m 4

h 14 k 22

h 7 k 11

81 a 12 b 6 c 38

9 a 6 b 3 c 19

144 x 4 y 80 ( b + 5 ) 16

12 x 2 y 40 ( b + 5 ) 8

w 5

w 2 w

w 7 z 3 k 13

w 3 z k 6 w z k

27 a 3 b 4 c 5 d 6

3 a b 2 c 2 d 3 3 a c

180 m 4 n 15 ( a 12 ) 15

6 m 2 n 7 ( a 12 ) 7 5 n ( a 12 )

Square roots involving fractions

A square root expression is in simplified form if there are

  1. no perfect squares in the radicand,
  2. no fractions in the radicand, or
  3. 3. no square root expressions in the denominator.

The square root expressions 5 a , 4 3 x y 5 , and 11 m 2 n a 4 2 x 2 are in simplified form.

The square root expressions 3 x 8 , 4 a 4 b 3 5 , and 2 y 3 x are not in simplified form.

Simplifying square roots with fractions

To simplify the square root expression x y ,

  1. Write the expression as x y using the rule x y = x y .
  2. Multiply the fraction by 1 in the form of y y .
  3. Simplify the remaining fraction, x y y .

Rationalizing the denominator

The process involved in step 2 is called rationalizing the denominator. This process removes square root expressions from the denominator using the fact that ( y ) ( y ) = y .

Sample set b

Simplify each square root.

9 25 = 9 25 = 3 5

3 5 = 3 5 = 3 5 · 5 5 = 15 5

9 8 = 9 8 = 9 8 · 8 8 = 3 8 8 = 3 4 · 2 8 = 3 4 2 8 = 3 · 2 2 8 = 3 2 4

k 2 m 3 = k 2 m 3 = k m 3 = k m 2 m = k m 2 m = k m m = k m m · m m = k m m m m = k m m · m = k m m 2

x 2 8 x + 16 = ( x 4 ) 2 = x 4

Practice set b

Simplify each square root.

81 25

9 5

2 7

14 7

4 5

2 5 5

10 4

10 2

9 4

3 2

a 3 6

a 6 a 6

y 4 x 3

y 2 x x 2

32 a 5 b 7

4 a 2 2 a b b 4

( x + 9 ) 2

x + 9

x 2 + 14 x + 49

x + 7

Exercises

For the following problems, simplify each of the radical expressions.

8 b 2

2 b 2

20 a 2

24 x 4

2 x 2 6

27 y 6

a 5

a 2 a

m 7

x 11

x 5 x

y 17

36 n 9

6 n 4 n

49 x 13

100 x 5 y 11

10 x 2 y 5 x y

64 a 7 b 3

5 16 m 6 n 7

20 m 3 n 3 n

8 9 a 4 b 11

3 16 x 3

12 x x

8 25 y 3

12 a 4

2 a 2 3

32 m 8

32 x 7

4 x 3 2 x

12 y 13

50 a 3 b 9

5 a b 4 2 a b

48 p 11 q 5

4 18 a 5 b 17

12 a 2 b 8 2 a b

8 108 x 21 y 3

4 75 a 4 b 6

20 a 2 b 3 3

6 72 x 2 y 4 z 10

b 12

b 6

c 18

a 2 b 2 c 2

a b c

4 x 2 y 2 z 2

9 a 2 b 3

3 a b b

16 x 4 y 5

m 6 n 8 p 12 q 20

m 3 n 4 p 6 q 10

r 2

p 2

p

1 4

1 16

1 4

4 25

9 49

3 7

5 8 3

2 32 3

8 6 3

5 6

2 7

14 7

3 10

4 3

2 3 3

2 5

3 10

30 10

16 a 2 5

24 a 5 7

2 a 2 42 a 7

72 x 2 y 3 5

2 a

2 a a

5 b

6 x 3

6 x x 2

12 y 5

49 x 2 y 5 z 9 25 a 3 b 11

7 x y 2 z 4 a b y z 5 a 2 b 6

27 x 6 y 15 3 3 x 3 y 5

( b + 2 ) 4

( b + 2 ) 2

( a 7 ) 8

( x + 2 ) 6

( x + 2 ) 3

( x + 2 ) 2 ( x + 1 ) 2

( a 3 ) 4 ( a 1 ) 2

( a 3 ) 2 ( a 1 )

( b + 7 ) 8 ( b 7 ) 6

a 2 10 a + 25

( a 5 )

b 2 + 6 b + 9

( a 2 2 a + 1 ) 4

( a 1 ) 4

( x 2 + 2 x + 1 ) 12

Exercises for review

( [link] ) Solve the inequality 3 ( a + 2 ) 2 ( 3 a + 4 )

a 2 3

( [link] ) Graph the inequality 6 x 5 ( x + 1 ) 6.
A horizontal line with arrows on both ends.

( [link] ) Supply the missing words. When looking at a graph from left-to-right, lines with _______ slope rise, while lines with __________ slope fall.

positive; negative

( [link] ) Simplify the complex fraction 5 + 1 x 5 1 x .

( [link] ) Simplify 121 x 4 w 6 z 8 by removing the radical sign.

11 x 2 w 3 z 4

Questions & Answers

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SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
types of nano material
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
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what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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Prasenjit
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Basic mathematics review. OpenStax CNX. Jun 06, 2012 Download for free at http://cnx.org/content/col11427/1.2
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