take any two digit
Eg: 55 square= 1. 5 square 25 and take 5 and carry down 2.
2. then multiply 5*5*2 u will get 50 and then take the carry down value2 now the total is 50+2=52.
3. Again take 2 and carry down 5.
4.now do the square of first digit and add the carry down number i.e.5.
Now the Answer is 3025.

kprat90 wrote:take any two digit
Eg: 55 square= 1. 5 square 25 and take 5 and carry down 2.
2. then multiply 5*5*2 u will get 50 and then take the carry down value2 now the total is 50+2=52.
3. Again take 2 and carry down 5.
4.now do the square of first digit and add the carry down number i.e.5.
Now the Answer is 3025.

Isn't much easier this way?

In squaring a two-digit number ending in 5, simply put 25 as the last two digit (of your answer) and multiply the digit in the ten's decimal place by a number next to it higher by 1...

that is, 5 x 6 (6 is a number next to 5 higher by 1)= 30

= 30'25

Here's another example... squaring 85

_ _ ' 2 5 (as the last digit of your answer)

8 x 9 = 72

Answer: 72'25

a certain method for squaring of numbers exists
as in if i wantd to figure out da answer for 33^2,the ans can be found by 33^2=35^2-34*4
if i used the same method for 34^2 naturally it would be 35^2-34.5*4
but my answer is incorrect. Is this method appropriate??

a certain method for squaring of numbers exists
as in if i wantd to figure out da answer for 33^2,the ans can be found by 33^2=35^2-34*4
if i used the same method for 34^2 naturally it would be 35^2-34.5*4
but my answer is incorrect. Is this method appropriate??

beat da gmat wrote:a certain method for squaring of numbers exists
as in if i wantd to figure out da answer for 33^2,the ans can be found by 33^2=35^2-34*4
if i used the same method for 34^2 naturally it would be 35^2-34.5*4
but my answer is incorrect. Is this method appropriate??

35^2 = 12'25 (can be done mentally)
but why the need to multiply 34 by 4?

The simplest way is by the use of this popular equation, (x+y)^2 - x^2 + 2xy + y^2

33^2 = 09'09 (Partial Square Value w/c is equivalent to x^2 + y^2)
30x3x2= 1 80 (Sub-product) w/c is equivalent to 2xy...
Total = 10'89
(of course with the conditions that x = 30 and y = 3)

w/ SSQ Method (Synthetic Squaring Method), the appropriate 'index squares' fall in their respective decimal places, such as...

34^2 = 09'16
3x4x2 = 2 4
Total= 11'56

Why complicate your self?

k wen a postive integer is given u have 2 take a postive number wile fining square roots n wen a odd no.is give u have to take an odd no
n it shod b a prime number

Please write in proper english language. Also, please explain clearly what you wish to tell. thanks

this method is excellent.... it helped me

we can find the sqare root by two methods
1)prime factorisation
2)Division method

PRIME FACTORISATION
Q)Find the square of 9801

3|9801
3|3267
3|1089
3|363
11|121
|11
prime factors are=3x3x3x3x11x11
make the pairs of aa
3x3x11=99
thus the square root of9801 is 99

IT IS A EASIEAR WAY TO FIND

to find square root of 9801 by division method
1) divide it into pairs (98)(01)
2) divide first pair with square of number and take that number as quotient and bring next pair with remainder 9 |(98)(01)| 9
______________________________ -81
_______________________________1701
now taking double of quotient divide again byng nearest number next to it such that by multiplying final number by number obtained the product is less than equal to dividend 189|1701|99
_____________________________1701
_____________________________0

this is somewhat difficult....

flixblixclix wrote:we can find the sqare root by two methods
1)prime factorisation
2)Division method

PRIME FACTORISATION
Q)Find the square of 9801...

Sorry, I rather prefer the easier method of:
1) Determining the first two digits
2) Determining the last digit
With the condition: THE GIVEN NUMBER IS A PERFECT SQUARE

OK, let's assume 9801 is a perk (perfect square number)
1) the first digit is 98
2) the last digit is 1

There is a technique w/c I called "Ten-Square Complements" in which there are always a pair of single digit numbers, when added will give us a sum of 10 and which their 'index squares' end, with the same digit number, such:

1 + 9 = 10, square of 1 = 01 while square of 9 = 81
2 + 8 = 10, square of 2 = 04 while square of 8 = 64
3 + 7 = 10, square of 3 = 09 while square of 7 = 49
4 + 6 = 10, square of 4 = 16 while square of 6 = 36
(Except 0 and 5, which don't have a pair)

9801 is a 4 digit number,
1) Regroup by twos - 98'01
2) Consider the first two digit, 98
The nearest square but less than 98 is 81 and its equivalent square root value is 9
3) The last digit of 98'01 is, of course, 1
There is a pair of single digit numbers with squares ending in 1 and these are 1 and 9
4) So,we have two possible square roots 91 and 98 and only one of them is correct

To determine which of the two is the right answer, simply consider the first digit of the possible answers (91 or 98) and "add 1" (that is, 9 + 1 = 10) and multiply to the original digit (9)
9 x 10 = 90

Conditions:
1) If the product (90) is less than the first two digits of the given problem (98), consider the much higher value among the two possible answers (between 91 and 98, choose 98)
2) If the product (90) is greater than the first two digits of the given problem (98),(which is not true or contradicting, in this case), consider the much lower value among the two possible answers (between 91 and 98, choose 91)

In this case, 90 is less than 98, therefore, (with the assumption that 9801 is a perk), the right answer would be: 98

NOTE: To make condition 2 agreeable, the given problem must be = 8,281
...2) If the product (90) is greater than the first two digits of the given problem (82), consider the much lower value among the two possible answers (between 91 and 98, choose 91)

Both prime factorization and division method were both traditionals and we're familiar with them.
These "alternative methods", go beyond the traditionals, or simply stated as "breaking the conventional strict rules we learned in the class rooms"

kalakita....chance ae illa

how would u noe the closest square number in a small amount of time

in my method i recommend prime factorization.....
for example to find √625

5|625
5|125
5|25
5|5
|1


Now find √5*5*5*5 = 5*5 = 25

hope dat ws easy..☺
---------------------------------------------
QED (Quite Easily Done)

to find 55² we can split the numbers as 5|5, (units place)²
multiply the tens digit by n*(n+1)
that is, 5 x 6 (6 is a number next to 5 higher by 1)= 30
= 30'25

How 2 find √39601 .............?? tell a quicker method plzzzzzzzz..........

shaurya.lawcitizen wrote:How 2 find √39601 .............?? tell a quicker method plzzzzzzzz..........

Try MSM-3 Method (Please read this thread from the beginning and you'll come up w/ the topic about MSM-3). This avoid repeating the explanations about MSM-3 or search in google " Square Root: A Digit Per Digit Approach by F. Mendoza, it would help you a lot). Thanks