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Standart 1.0. Finding Sguare Roots

1.0. FINDING SGUARE ROOTS

1.1. THE GENERAL METHOD


Look at the multiplication tab :

1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
6 12 18 24 30 36 42 48 54 60
7 14 21 28 35 42 49 56 63 70
8 16 24 32 40 48 56 64 72 80
9 18 27 36 45 54 63 72 81 90
10 20 30 40 50 60 70 80 90 100



The diagonal from the top left corner to the bottom right corner shows the numbers obtained when we square the positive integers from 1 to 10. we call the elements of the set ( 1,4,9,16,25,36,49,64,81,100… )
Square numbers.
Square numbers are the squares of the integers and thus have integral square roots ( square roots that are integers ).

√1 = 1, √4 = 2, √9 = 3, √16 = 4, √25 = 5,

√36 = 6, √49 = 7, √81 = 9, √100 = 10 …

If n is an integer then n² is a square number and we can write the square root as:

√ n² = n

In the same way the cubic numbers, or elements of the set :
{ 1, 8, 27, 64, 125, 256, 343 … }

are the cubes of the integers and thus have integral cubes roots.

∛ 1 = 1, ∛ 8 = 2, ∛ 27 = 3, ∛ 64 = 4, ∛ 125 = 5,


and so on, and if n is an integer then n³ is a cubic number and we can write the cubed root as :

∛ n³ = n

Notice that :

1 x 1 = 1 ⇒ √1 = 1 ⇒√10⁰ = 10⁰
10 x 10 = 100 ⇒ √100 = 10 ⇒√10² = 10¹
100 x 100 = 10000 ⇒ √10000 = 100 ⇒√10⁴ = 10²
1000 x 1000 = 1000000 ⇒ √1000000 = 1000 ⇒√10⁶ = 10³
and so on.

We see that the square root of 10 raised to an even power is 120 raised to a half of that power.
Obviously, odd powers of 10 do not have integral square roots.
For example, we know that √9 = 3 and √16 = 4. So √10 must be an irrational number between 3 and 4.

√9 < √10 < √16
⇒ 3 < √10 < 4


The general method for finding a square root.

Example: Find √784


2 8
7
4
84

84
84
3
3
0


Here are the steps:

1. Separate the hundreds digit ( 7 ) from the tens and units digits ( 84 ) of the number.

2. Since: √4 < √7 < √9
Then : 2 < √7 < 3

Which means that we can write:

20 < √700 < 30
and the tens digit of √784 must be 2
Write 2 above 7 and on the left of 7.
3. Write 4, the square of 2, under 7 and subtract it from 7.
write the difference ( 3 ) under the 4.
4. Bring down the next group of digits ( 84 ) and write them next to the 3 to give us 384.
5. Double the first part of the result ( 2 ) and write 4 to the left of 384 leaving a space for another number.
6. The units digit of the result, N, w,ill now be written above 84 and in the space to the right of 4 such that :

N x 4N ≤ 384

The quotient when 38 is divided by 4 is 9, so we should try 9 first.
However, since 9 x 49 = 441, which is bigger than 384, we now try 8.
8 x 48 = 384 so the units digit is 8 and we can write:

√784 = 28

Which we can check by squaring 28.


1.2. THE METHOD OF FACTORS
Real numbers that are written using root symbols are called radicals.
√2, √3, √4, √5, ∛4, ∛8, ∜9 and ∜16 are example of radicals.
An irrational number is written using a root symbol then we can also call it a surd.
Of the above list of radicals √2, √3, √5, ∛4 and ∜9 are surds.
Since the radicals √4, ∛8 and ∜16 have integral roots then they are rational numbers. So they are not surds.
The irrational number π is not a surd since it is not a radical.
Suppose we take a square number, find its square root and then square it. Obviously, we come back to the original square number.
For example, Z36 = 6 and 6² = 36
So we can write: ( √36 ) ² = 36
The operations of finding the square root of a number and squaring a number are clearly inverse operations.
Although we cannot find irrational square roots exactly we know that the square of the root must bring us back to the original number.

For example: √2 = 1.414… and ( 1.414…) ² = 2
Or: ( √2 ) ² = 2

And if a is any non – negative integer then :





We call this the multiplication property of roots.
If a, b ε Z ⁺ then:






We can check this using a square number that contains two or more other square numbers as its factors. For example :

√36 = √9 x 4 = √9 x √4 = 3 x 2 = 6

Which is true since 6² = 36

Obviously, the multiplication property of roots can be applied to any number of factors:
√36 = √4 x 4 x 4 = √4 x √4 x √4 = 2 x 2 x 2 = 8

and 8² = 64

in fact, we can use the multiplication property of roots to find the square root of any square number that is not a prime number. We call this the method of factors for finding a square root. The method depends on the fact that, if we take any positive integer and square it, the resulting square number must contain two of each of the prime factors of the original number.



Example : Simplify √144

We begin by factorising 144

144 2
72 2
36 2
18 2 144 = 2. 2. 2. 2. 3. 3.
9 3
3 3
1
so √144 = √2.2.2.2.3.3 = √2² . 2² . 3²
= √2² x √2² x √3²
= 2 x 2 x 3
= 12
so √144 = 12
ıt is clear from this example that √144 is a rational number since its prime factors can be arranger in pairs with no factors left over.
Simplifying surds to the form a √b
We can use the method of factors to simplify irrational square roots to their simplest form.
Example2: Simplify Z180

180 2
90 2
45 3
15 3 180 = 2. 2. 3. 3. 5
5 5
1
so √180 = √2.2.3.3.5 = √2². 3². 5
= √2² x √3² x √5
= 2 x 3 x √5
= 6√5
so √180 = 6 √5 in its simplest form.
Example: Simplify 8 √12

12 2
6 2 √12 = √2.2.3 = √2² . 3 = 2√3
3 3
1 ⇒ 8 √12 = 8 x 2√3 = 16 √3

Expressing surds as the root of a single number
We have seen that in general:
√a²b = √a² x √b = a x √b = a √b
if we reserve the process:
a √b = a x √b = √a² x √b = √a²b
We see that, to write an irrational radical as the root of a single number, we must square the number in front of the root and multiply it by the number inside the root.
For example, we have seen that √180 = 6 √5. So to write 6 √5 as the root of a single number we must put the number 6 back into the root by squaring it:
6 √5 = √6² x √5 = √6².5 = √36.5 = √180
this process is equivalent to checking a square root that is rational by squaring it back.
For example, to check that √144 = 12 we write:
12 = √12² = √144
2.0. THE ADDITION AND SUBTRACTION OF RADİCALS
2.1. Radicals having common roots
we can add and subtract radicals only if they have a common root. To do this we use the distributive property of multiplication over addition of subtraction with the root part as the common factor.
Example: Simplify 2√6 + 3√6
2√6 + 3√6 = ( 2 + 3 ) √6 = 5√6
ıt is important to understand that √6 is a symbol for an irrational number in the same way that π is a symbol for an irrational number. If we wanted to find the total area of two circles, one having an area of 2 π cm² and the other having an area of 3 π cm² then we would use a similar method to find their sum.
2 π + 3 π = ( 2 + 3 ) π = 5 π cm²
so, in the same way that x + x = 2x and π + π = 2 π, we have:
√6 + √6 = 1 √6 + 1 √6 = ( 1 + 1 ) √6 = 2 √6
Example: Simplify √2 + 2√2 - 5√2
√2 + 2√2 - 5√2 = ( 1 + 2 – 5 ) √2 = - 2√2
if two or more radicals simplify to a common root then we can add or subtract them.
Example: Simplify √12 - √27 + 2√48
√12 - √27 + 2√48
= √4.3 - √9.3. + 2√16.3
= 2√3 - 3√3 + 2. 4√3
= 2√3 - 3√3 + 8√3
= ( 2 - 3 + 8 ) √3
= 7 √3
2.2. Radicals having different roots
we cannot add or subtract radicals if their root parts are different since. We cannot use the distributive property.
The result of adding √2 to √3 is written simply
√2 + √3
in the same way that we write the sum of x and y as:
x + y
however, if any radicals in an expression do have common roots then we must simplify them.
Example: Simplify √32 - √54 - 2√8 + √24
√32 - √54 - √8 + √24
= √16.2 - √9.6 - √4.2 + √4.6
= 4√2 - 3√6 - 2√2 + 2√6
= 4√2 - 2√2 - 3√6 + 2√6
= ( 4 – 2 ) √2 + ( -3 + 2 ) √6
= 2√2 + ( -1 ) √6
= 2√2 - √6
We simplify the radicals containing √2 and the radicals containing √6 separately and the result consists of an expression consisting of both radicals.
3.0. THE MULTIPLICATION AND DIVISION OF RADICALS
3.1. The multiplication of radicals
We know that multiplication is a quick way of doing repeated additions of the same number.
So if: √2 + √2 + √2 = ( 1 + 1 + 1 ) √2 = 3√2
Then: 3 x √2 = 3√2
And we see that to multiply a rational number with a surd we simply write the rational number in front of the surd.
We call the number in front of a root in a radical the coefficient of the radical. So in the radical 3√2 the number 3 is the coefficient.
Now if : 2√5 + 2√5 + 2√5 + 2√5 = ( 2 + 2 + 2 + 2 ) √5 = 8√5
Then : 4 x 2√5 = 8√5
And we see that to multiply a rational number with a radical we multiply the rational number with the coefficient of the radical to obtain the coefficient of the result.
The multiplication property of roots:
Shows us how to multiply a root by another root.

If the roots are the same number then we have:
√a x √a = √a x a = √a² = a
For example: √3 x √3 = √3 x 3 = √3² = 3
Clearly, the product of a root with itself always gives a rational number as the result. To square the square root of a number must bring us back to the number.
The results so far show us how to multiply a radical by another radical having the same root.
Example: 4 √6 x 2 √6 = 4 x √6 x 2 x √6 = 4 x 2 x √6 x √6
= 8 x √6² = 8 x 6 = 48
you can see that we multiply the coefficients of the radicals together and we multiply the root parts together. We then multiply these two results together to obtain the final product.
Notice that, since the root parts are the same, the result is again a rational number.
If we apply the multiplication property of roots to two different roots then the result is always an irrational number.
Examples : √2 x √3 = √2 x 3 = √6
And : 4 √2 x 5 √3 = 4 x 5 x √2 x √3 = 20 √6
After performing an operation on radicals we should always simplify the root part if possible.
Example : Simplify 3√6 x 4√2
3√6 x 4√2 = 3 x 4 x ( √6 x √2 )
= 12 x √12
= 12 x 2√3
= ( 12 x 2 ) x √3
= 24 x √3
= 24 √3
3.2. Expansions of radical expressions
we have already seen how the distributive property of multiplication over addition lets us write:
a ( b + c ) = ab + ac
If a , b and c are real numbers we can show this fact in the following diagram:





Where a and ( b + c ) = ab + ac
Gives the area of the rectangle.






The diagram above shows:
√2 ( 3 + √6 ) = 3√2 + √12 = 3√2 + 2√3
When we use the distributive property to remove parentheses in this way, we call it expanding the expression.
We call an expression consisting of a single term, like √2, a monomial.
An expression consisting of two terms, like ( 3 + √6 ), is called a binomial.
Later we shall be looking at expressions consisting of three terms, like ( √2 + √3 + 2 ). We call these trinomials.
In the above example we have the expansion of a monomial with a biominal. Here is an example with two binomials.
√6
4√2

3√3

12






The area of the rectangle is : ( √2 + 3 ) ( √3 + 4 )
= √2 ( √3 + 4 ) + 3 ( √3 + 4 ) = √6 + 4√2 + 3√3 + 12
You can see that we multiply the second binomial by √2 and then by 3.
Here is an example of an expansion involving the product of a binomial with a trinomial.

Expand and simplify: ( 3√2 - 2√3 ) ( 2√2 - 3√3 + √6 )
( 3√2 - 2√3 ) ( 2√2 - 3√3 + √6 )
= 3√2 ( 2√2 - 3√3 + √6 ) - 2√3 ( 2√2 - 3√3 + √6 )
= 6√4 - 9√6 + 3√12 - 4√6 + 6√9 - 2√18
= 6.2 – 9√6 + 3.2 √3 - 4√6 + 6.3 – 2.3 √2
= 12 - 9√6 + 6√3 - 4√6 + 18 - 6√2
= 12 + 18 - 9√6 - 4√6 + 6√3 - 6√2
= 30 - 13√6 + 6√3 - 6√2
3.3. Rationalizing monomial denominators
For example, √2 and 1 actually represent the same number. However, √2
2 √2 2
is written with a rational denominator while 1 has an irrational denominator.
√2
mathematicians and scientists have decided that the standard form of such fractions should have a rational denominator.
The process of changing the denominator of an irrational fraction from an irrational number to a rational number is called rationalizing the denominator.
Rationalizing the denominator of a fraction having a monomial radical denominator depends on the fact that the square of a square root is a rational number. So, by multiplying the numerator and the denominator by the root part of the denominator, we obtain the standard form of the radical.
Example : Rationalize the denominator of 1
√2
1 = 1 x √2 = √2 = √2
√2 √2 x √2 (√2)² 2

Example : Rationalize the denominator of 3√2
2 √3
3√2 = 3√2 x √3 = 3√6 = √6
2√3 2√3 x √3 2.3 2
Division of radicals
In the same way that we use the multiplication property of roots to find the product of roots :
√a x √b = √ab
so we use the division property of roots to find the quotient of two roots



For example, since √4 x √9 = √4.9 = √36
So, √36 = √36 / 4 = √9 = 3
√4
Which is true since :
√36 = 6 = 3
√4 2
Also, √36 = 36 = √4 = 2
√9 9
And, √36 = 6 = 2
√9 3
Example : Simplify 4
5
4 = √4 = 2 = 2 x √5 = 2√5
5 √5 √5 √5 x √5 5
The division property of roots is applied the denominator of the result is then rationalized.
Suppose we use the division property of roots to simplify √6
√2
√6 = 6 = √3
√2 2
You can see that since 6 ÷ 2 = 3 so:
√6 ÷ √2 = √3
and this means that we can cancel a root with a root in the same way that we can cancel an integer with an integer.
√3
√6 = √3 = √3
√2 1
√1
Example : Simplify √28 + √112
√350 - √126
√28 + √112 = 2√7 + 4√7 = 6√7 = 3 = 3√2
√350 - √126 5√14 - 3√14 2√7 √2 2
3.4. Rationalizing Binomial Denominators
If we expand ( √3 + √2 ) ( √3 - √2 ) we get:
( 3 √2 ) ( √3 - √2 = 3 - √6 + √6 – 2 = 1
and the expansion of ( 3√6 - 4√3 ) ( 3√6 + 4√3 ) gives us:
( 3√6 - 4√3 ) ( 3√6 + 4√3 ) = (3√6) ² + (3√6) (4√3) – (4√3) (3√6) – (4√3)²
= 54 – 48
= 6

Notice that in each of these examples the results are rational numbers, 1 and 6. In general, if we expand a pair of binomials, or expressions with two terms, of the form
( √a + √b ) ( √a - √b ) where a and b are rational numbers we obtain:
( √a + √b ) ( √a - √b ) = a - √ab + √ab – b = a – b
which is a rational number since the two irrational parts of the expansion, - √ab and √ab, sum to zero.
We call a pair of binominals of the form (√a + √b) (√a - √b) conjugates.
So ( √3 + √2 ) and ( √3 - √2 ) are a pair of conjugates.
Also ( 3√6 - 4√3 ) and ( 3√6 + 4√3 ) are a pair of conjugates.
We say that ( √3 + √2 ) is the conjugate of ( √3 - √2 ) and that ( √3 - √2 ) is the conjugate of ( √3 + √2 ).
Also ( 3√6 - 4√3 ) and ( 3√6 + 4√3 ) are conjugates of each other.
We can use the fact that, when we expand a pair of conjugates of the form (√a + √b ) ( √a - √b ) we always obtain a rational number, to rationalize the denominators of fractions having irrational binomial denominators. By multiplying both the numerators and denominators of the fraction by the conjugate of the denominator we obtain a rational denominator.

Example : Rationalize the denominator and simplify:
√6 + 3
√3 + √2
Multiplying the numerator and denominator by the conjugate of the denominator:
( √6 + 3 ) ( √3 - √2 )
( √3 + √2 ) ( √3 - √2 )

= 3√2 - 2√3 + 3√3 - 3√2
3 -2
= √3 = √3
1
QUESTIONS ABOUT RADICALS BETWEEN 1990 – 2000 FEN LİSESİ EXAMS
Questions 1: ( 1990 )

1 + 1 + 9 = simplify :
16

= 1 + 25
16

= 1 + 5
4
= 9
4
= 3 / 2
Questions 2 : ( 1991 )
a,b,c different even numbers if √ab . √ac . √bc = m , find m ?
a² b² c²
= a.b.a.c.b.c
a² b² c²
= a.b.c
a² b² c²
= 1
abc

Questions 3 : ( 1991 )
√0,16 + √2,25 + √1,21 = a , find a ?
√0,25 + √1,44 √2,89

a = 0,4 + 1,5 + 1,1
0,5 + 1,2 1,7
a = 1,9 + 1,1
1,7 1,7
a = 3
1,7
a = 30
17

Questions 4 : ( 1992 )
√1,28 - √2,42 + √0,32 simplify :
= √ (0,64).2 - √(1,21).2 + √(0,16).2
= (0.8).√2 – (1,1).√2 + (0.4).√2
= (0,1)√2
= √2
10
Questions 5 : ( 1993 )
√12 , √3 find the geometric mean of them ?
= √12 . √3

= √36 = √6

Questions 6 : ( 1993 )
0,6 simplify:
√0,9 . √0,4
= 0,6
√0,36
= 0,6 = 1
0,6

Questions 7 : ( 1994 )
10. ( √0,27 + √0,48 - √1,08 ) : 1 simplify:
√3
= 10. (0,3. √3 + 0,4. √3 - 0,6. √3). √3
= 10. ( 0,1. √3 ) . √3
= 3

Questions 8 : ( 1995 )
√121 - 8² + 6² simplify:
√5 . √20

= 11 - √100 = 11 - 10 = 1
√100 10




Questions 9 : (1995 )

(√2-√3)² + √24 simplify :
√25
= ( 2 – 2.√2.√3 + 3 ) + 2√6
5
= 2 - 2√6 + 3 + 2√6
5
= 5
5
= 1


Questions 10 : (1995 )
If a = √2 , b = √3 , c = √12, c³ = ?
a² b
= c³ = ( √12 ) ³
a² b ( √2 ) ² . √3

= 12 √12
2 √3

= 12 √12
√1 2

= 12










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