ESTIMATE SQUARE ROOTS
When we estimate the square root of a number, first we will find the two values between which the square root of the given number lies.
For example square root of the number 40 lies between 6 and 7.
Because square root of 36 is 6 and square root of 49 is 7. The given number lies between 36 and 49. So that we can say that the square root of the given number is between 6 and 7.
√1 = 1
√4 = 2
√9 = 3
√16 = 4
√25 = 5
√36 = 6
√49 = 7
√64 = 8
√81 = 9
√100 = 10
√121 = 11
√144 = 12
√169 = 13
√196 = 14
√225 = 15
√256 = 16
√289 = 17
√324 = 18
√361 = 19
√400 = 20
√441 = 21
√484 = 22
√529 = 23
√576 = 24
√625 = 25
√676 = 26
√729 = 27
√784 = 28
√841 = 29
√900 = 30
From the above square roots, we can come to know that
(i) If a perfect square has ‘n’ digits where n is even, its square root has n/2 digits.
(ii) If a perfect square has ‘n’ digits where n is odd, its square root has (n + 1)/2 digits.
Estimate the values of the following square roots to the nearest whole number :
Example 1 :
√80
Solution :
80 is not a perfect square.
Find the two perfect squares surrounding 80.
They are 64 and 81.
Then, we have
64 < 80 < 81
Taking square root,
√64 < √80 < √81
8 < √80 < 9
The value of √80 lies between 8 and 9.
In the inequality 64 < 80 < 81, since 80 is closer to 81, the approximate value of √80 is 9.
Example 2 :
√1000
Solution :
1000 is not a perfect square.
Find the two perfect squares surrounding 1000.
They are 961 and 1024.
Then, we have
961 < 1000 < 1024
Taking square root,
√961 < √1000 < √1024
31 < √1000 < 32
The value of √1000 lies between 31 and 32.
In the inequality 961 < 1000 < 1024, since 1000 is closer to 1024, the approximate value of √1000 is 32.
Example 3 :
√172
Solution :
172 is not a perfect square.
Find the two perfect squares surrounding 172.
They are 169 and 196.
Then, we have
169 < 172 < 196
Taking square root,
√169 < √172 < √196
13 < √172 < 14
The value of √172 lies between 13 and 14.
In the inequality 169 < 172 < 196, since 172 is closer to 169, the approximate value of √172 is 13.
Example 4 :
√5928
Solution :
5928 is not a perfect square.
Find the two perfect squares surrounding 5928.
They are 5776 and 5929.
Then, we have
5776 < 5928 < 5929
Taking square root,
√5776 < √5928 < √5929
76 < √5928 < 77
The value of √5928 lies between 76 and 77.
In the inequality 5776 < 5928 < 5929, since 5928 is closer to 5929, the approximate value of √5928 is 77.
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