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Squares and Square Roots

    Finding the Square of a Number

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8th class > Squares and Square Roots > Finding the Square of a Number

Finding the Square of a Number

Squares of small numbers like 3, 4, 5, 6, 7, ... etc. are easy to find. But can we find the square of 23 so quickly?

The answer is not so easy and we may need to multiply 23 by 23.

There is a way to find this without having to multiply 23 × 23.

We know

23 = 20 + 3

Therefore

232 = 20+32= 20(20 + 3) + 3(20 + 3)

= 202 + 20 × 3 + 3 × 20 + 32

= + + + =

Example 1: Find the square of the following numbers without actual multiplication.

(i) 392 = 30+92 = 30(30 + 9) + 9(30 + 9)

= 302 + 30 × 9 + 9 × 30 + 92

= + + + =

(ii) 422 = 40+22 = 40(40 + 2) + 2(40 + 2)

= 402 + 40 × 2 + 2 × 40 + 22

= + + + =

Other patterns in squares

Consider the following pattern:

252 = 625 = (2 × 3) hundreds + 25

352 = 1225 = (3 × 4) hundreds + 25

752 = 5625 = (7 × 8) hundreds + 25

1252 = 15625 = (12 × 13) hundreds + 25

952 = 9025 = ( × ) hundreds + 25

Consider a number with unit digit 5, i.e., a5

a52=10a+52

= 10a10a+5+510a+5

= 100a2+50a+50a+25

= 100aa+1+25

= aa+1hundred+25

Try these

Find the squares of the following numbers containing 5 in unit’s place:

(i) 15

152 = 100×1×1+1+25 = 100×2+25 = + =

(ii) 95

952= 10×9+52 = 100×9×9+1+25 = 100×90+25 = + =

(iii) 105

1052 = 10×10+52 = 100×10×10+1+25 = 100×110+25 = + =

(iv) 205

2052 = 10×20+52 = 100×20×20+1+25 = 100×420+25 = + =

Pythagorean triplets

Consider the following

32 +42 = + =

= 52

The collection of numbers 3, 4 and 5 is known as Pythagorean triplet.

We also have

62 +82 = + =

= 102

making 6, 8 and 10 another example of such triplet.

Similarly,

52 + 122 = + =

=132.

The numbers 5, 12, 13 form another such triplet.

For any natural number m > 1, we have:

2m2+m2−12=m2+12.

So, 2m,m2 – 1 and m2 + 1 forms a Pythagorean triplet.

Example 2: Now, write a Pythagorean triplet whose smallest member is 8 using the above general form

Instructions

The smallest triplet is 8

  • Say, m2 – 1 = 8
  • Upon solving we get, m =
  • Thus, m2+ 1 = and 2m =
  • This gives us the Pythagorean triplet of 6,8,10. However, as it turns out 8 isn't the smallest member. So, let's try another assumption.
  • Let's take 2m = 8
  • Which gives m =
  • Thus, m2+ 1 = and m2- 1 =
  • We have found the pythagoras triplet of 8,15 and 17 with 8 as the smallest member.

Example 3: Find a Pythagorean triplet in which one member is 12

Instructions

One triplet member is 12

  • Say, m2 – 1 = 12
  • Upon solving we find that: m isn't an integer. Let's try another assumption.
  • If we try m2 + 1 = 12, we get m2 = 11 which further gives a non-integer value for m
  • Thus, we can only take 2m = which gives m =.
  • Thus, m2+ 1 = and m2- 1 =
  • We have found the pythagoras triplet of 12,35 and 37.

This shows that our assumption when trying to find a suitable value of 'm' is important.

Note: All Pythagorean triplets may not be obtained using this form. For example: another triplet 5, 12, 13 also has 12 as a member.