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Perfect Square Number
A perfect square number is the product of two integers among the whole number. It is the perfect root of a particular number. For example:
1× 1 = 1^{2}(1 squared) = 1 (1 is a perfect square number)
2× 2 = 2^{2}(2 squared) = 4 (4 is a perfect square number)
3× 3 = 3^{2} (3 squared) = 9 (9 is a perfect square number)
Thus, a perfect square number is the product of two identical number 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 . . . . . . . . . . . and so on.
Square Root
A number that creates a specified number when it is multiplied by itself is said to be a square root. It is the product of two equal number. The radical symbol (\(\sqrt{}\)) is used to denote square root of a number. For example:
\(\sqrt{1}\) = 1
\(\sqrt{4}\) = 2
\(\sqrt{9}\) = 3
\(\sqrt{64}\) = 8
Note:
1 is the square number of 1, so 1 is called the square root of 1.
9 is the square number of 3, so 3 is called the square root of 9.
49 is the square number of 7, so 7 is called the square root of 49.
Factorization Method
In order to find out the square roots of a perfect square number, factorization method is used. For example:
Find the square root of 64 using factorization method.
So,
64 = 2× 2× 2× 2× 2× 2
= 2^{2}×2^{2}×2^{2}
∴ \(\sqrt{64}\) = 2× 2× 2 = 8
Division Method
In order to find out the square roots of decimals and larger numbers, division method is used. For example:
Find the square root of 2704 using division method.
There is one unit cube so 1 is a cubic number.
There are 8 unit cubes. So, 8 is a cubic number.
There are 27 unit cubes, so 27 is a cubic number.
1^{3} = 1× 1× 1 = 1 (1 is the cube of 1 & 1 is the cube root of 1)
2^{3} = 2× 2× 2 = 8 (8 is the cube of 2 & 2 is the cube root of 8)
5^{3} = 5× 5× 5 = 125(125 is the cube of 5 & 5 is the cube root of 125)
7^{3} = 7 × 7 × 7 = 343 (343 is the cube of 7& 7is the cube root of 343)
Cube Number:Cubic number is the product of three iderntical number.
Cube root of a cubic number:Cube root of a cubic number is one of the identical numbers.It is denoted by the symbol \(\sqrt[3]{}\). For example:
or, \(\sqrt[3]{1}\) = 1
or,\(\sqrt[3]{27}\) = 3
or,\(\sqrt[3]{729}\) = 9 etc.
The list of number or object in a specific order is said to be a sequence. For example:
Rule 1:
Lets supopose a sequence 2, 4, 6, 8, 10, 12, . . . . . .
Here, the common difference between each consecutive pair is 2. Consider 'n' as the number of terms of the sequence.
Then, the common difference is 2 and & the first term of the rule be 2n+ . . . . . . . . . . or 2n- . . . . . . . . . .
To get the first term 2,
n = 1
2n± . . . . . . .
= 2× 2 + 0 = 4
Similarly,
To get the second term 4,
n = 1
2n± . . . . . . .
= 2× 2 + 0 = 2
Here,
To get the n^{th} term o0f the sequence, the rule must be 2n. In this way, we can find out the n^{th} term of any sequence.
Rule 2:
Lets suppose the sequence 2, 5, 8, 11, 14, . . . . . . .
Consider that a denotes the first term and d denotes the common difference of the sequence.
d = 5 - 2 = 3 | d = 8 - 5 = 3 | d = 11 - 8 = 3 | d = 14 - 11 = 3 |
2 + 3 = 5 | 5 + 3 = 8 | 8 + 3 = 11 | 3 + 11 = 14 |
From the above illustration,
The 1^{st} term (t_{1}) = a = 2
The 2^{nd} term (t_{2}) = a + (2 - 1)d = a + d = a
The 3^{rd} term (t_{3}) = a + (3 - 1)d = a + 2d
The 4^{th} term (t_{4}) = a + (4 -1)d = a + 3d
∴ n^{th} term (t_{n}) = a + (n - 1)d
by using the n^{th} term, we can find the 12^{th} term of the sequence as :
8^{th} term (t_{8}) = a + (8 - 1)d
= a + 7d
= 2 + 7× 3
= 23
Solution:
144 ÷ 2 = 72 (remainder is 0)
72 ÷ 2 = 36 (remainder is 0)
36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (renainder is 0)
9 ÷ 3 = 3 (remainder is 0)
Now,
144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 2^{2} × 2^{2} × 3^{2 }
∴ \(\sqrt{144}\) = 2 × 2 × 3
= 12
Solution:
576 ÷ 2 = 288 (remainder is 0)
288 ÷ 2 = 144 (remainder is 0)
144 ÷ 2 = 72 (remainder is 0)
72 ÷ 2 = 36 (remainder is 0)
36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (remainder is 0)
Now,
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
576 = 2^{2} × 2^{2} × 2^{2} × 3^{2}
∴ \(\sqrt{576}\) = 2 × 2 × 2 × 3
= 24
Solution:
= \(\sqrt{8}\) × 3\(\sqrt{18}\) × 2\(\sqrt{48}\)
= \(\sqrt{4 × 2}\) × 3\(\sqrt{9 × 2}\) × 2\(\sqrt{16 × 3}\)
= 2\(\sqrt{2}\) × 3 × 3\(\sqrt{2}\) × 2 × 4\(\sqrt{3}\)
= 144\(\sqrt{4}\)
= 144\(\sqrt{4 × 3}\)
= 144 × 2\(\sqrt{3}\)
= 288\(\sqrt{3}\)
Solution:
1.96 = \(\frac{196}{100}\)
∴ \(\sqrt{1.96}\) = \(\sqrt{\frac{196}{100}}\)
= \(\frac{\sqrt{2 × 2 × 7 × 7}}{\sqrt{2 × 2 × 5 × 5}}\)
= \(\frac{\sqrt{2^2 × 7^2}}{\sqrt{2^2 × 5^2}}\)
= \(\frac{2 × 7}{2 × 5}\)
= \(\frac{14}{10}\)
= 1.4
Solution:
= \(\sqrt{\frac{256}{625}}\)
= \(\sqrt{\frac{2 × 2 × 2 × 2 × 2 × 2 × 2 × 2}{5 × 5 × 5 × 5}}\)
= \(\sqrt{\frac{2^2 × 2^2 × 2^2 × 2^2}{5^2 × 5^2}}\)
= \(\frac{2 × 2 × 2 × 2}{5 × 5}\)
= \(\frac{16}{25}\)
______ is the product of three identical number.
Square root is denoted by the symbol ______.
ASK ANY QUESTION ON Root and Sequence Number
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