‹ Class 8 · Ch 6
We Distribute, Yet Things Multiply · Principle 12 of 13

Quick squaring

a² = (a + b)(a − b) + b².

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NCERT: 6.2 Special Cases of the Distributive Property

Think

Square 31 the quick way

Squaring 31 takes a while. But 32 × 30 is easy: 32 × 30 = 960.

How is 31² related to 32 × 30?

What this lesson covers

The idea

Rewriting the difference of squares as a² = (a + b)(a – b) + b² lets a square be computed quickly by choosing b so that a + b and a – b are easy to multiply.

Square 31 the quick way

Squaring 31 takes a while. But 32 × 30 is easy: 32 × 30 = 960.

How is 31² related to 32 × 30?

  • They are equal
  • 31² is 1 more
  • 31² is 1 less

Choose b

The rule works for any b. Choose b so that one of the two numbers ends in 0. Then the multiplication is easy.

Pick b to make it easy

a² = (a + b)(a − b) + b². Choose b so that a + b and a − b are easy to multiply.

Why is it true? We know (a + b)(a − b) = a² − b² Add b² to both sides.

31² = (31 + 1)(31 − 1) + 1² = 32 × 30 + 1 = 961

197² = (197 + 3)(197 − 3) + 3² = 200 × 194 + 9 = 38809 Sridharacharya (750 CE) gave this method.

Notes

a² = (a + b)(a − b) + b²: choose b so that a + b and a − b are easy to multiply.

Check yourself

Find 98² with b = 2: (98 + 2)(98 − 2) + 2²

Answer: 9604

(98 + 2)(98 − 2) + 2² = 100 × 96 + 4 = 9600 + 4 = 9604.

Which b makes 83² easiest?

Find 203². Take b = 3, so that 203 − 3 = 200.

Answer: 41209

(203 + 3)(203 − 3) + 3² = 206 × 200 + 9 = 41200 + 9 = 41209.

Why is a² = (a + b)(a − b) + b² true?

  • b = 1: 84 × 82 + 1². Neither 84 nor 82 ends in 0, so 84 × 82 is not easy.
  • b = 3: 86 × 80 + 3² — correct. Yes! 80 is round: 86 × 80 + 9 = 6880 + 9 = 6889.
  • b = 5: 88 × 78 + 5². Neither 88 nor 78 ends in 0, so 88 × 78 is not easy.
  • (a + b)(a − b) = a² − b², so adding b² to both sides gives a² — correct. Yes! It follows from the difference of squares.
  • Because a + b and a − b are both close to a. Close or not, that is not the reason. It follows from (a + b)(a − b) = a² − b².
  • Only when b is small. It is true for every b. The identity is exact, whatever b is.
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