Ganita Prakash | Grade 8 Figure it Out Chapter-6 | We Distribute, Yet Things Multiply- Solution Page-149

Figure it Out 

1. Which is greater: (a – b) 2 or (b – a) 2 ? Justify your answer. 

2. Express 100 as the difference of two squares. 

3. Find 4062 , 722 , 1452 , 10972 , and 1242 using the identities you have learnt so far. 

4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.

SOLUTION:

1. Which is greater: (a − b)² or (b − a)²? Justify.

• Note that b − a = −(a − b).
• Squaring removes the sign: (b − a)² = [−(a − b)]² = (a − b)².
• Therefore, they are equal for all real numbers a and b.

Conclusion: (a − b)² and (b − a)² are equal.

2. Express 100 as the difference of two squares.
   A number N can be written as a difference of squares if N = x² − y² = (x − y)(x + y).
   We need (x − y)(x + y) = 100. Factor 100 and pair factors with the same parity (both even or both odd), so that x and y are integers:

• 100 = 100·1 → x = (100 + 1)/2 = 50.5, y = (100 − 1)/2 = 49.5 (not integers)
• 100 = 50·2 → x = (50 + 2)/2 = 26, y = (50 − 2)/2 = 24 → 26² − 24² = 100
• 100 = 25·4 → x = (25 + 4)/2 = 14.5, y = (25 − 4)/2 = 10.5 (not integers)
• 100 = 20·5 → x = (20 + 5)/2 = 12.5, y = (20 − 5)/2 = 7.5 (not integers)
• 100 = 10·10 → x = 10, y = 0 → 10² − 0² = 100

So, examples:

• 26² − 24² = 100
• 10² − 0² = 100

If you want all integer solutions: (x, y) = (k + 50/k)/2, (k − 50/k)/2 for any integer k dividing 100 with x, y integers, i.e., use factor pairs of 100 with same parity.

3. Find 406², 72², 145², 1097², and 124² using identities.
   We’ll use (x ± y)² = x² ± 2xy + y² and (x + y)(x − y) where convenient.

• 406² = (400 + 6)² = 400² + 2·400·6 + 6² = 160000 + 4800 + 36 = 164836

• 72² = (70 + 2)² = 70² + 2·70·2 + 2² = 4900 + 280 + 4 = 5184

• 145² = (100 + 45)² = 10000 + 2·100·45 + 45²
  = 10000 + 9000 + 2025 = 21025
  (Alternatively: (150 − 5)² = 22500 − 1500 + 25 = 21025)

• 1097² = (1100 − 3)² = 1100² − 2·1100·3 + 3²
  = 1,210,000 − 6,600 + 9 = 1,203,409

• 124² = (120 + 4)² = 14400 + 960 + 16 = 15376

Answers:

• 406² = 164,836
• 72² = 5,184
• 145² = 21,025
• 1097² = 1,203,409
• 124² = 15,376

4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify.
   I’ll interpret “Pattern 1 and 2” as standard identities commonly taught before these exercises:

• Pattern 1: (a + b)² = a² + 2ab + b²
• Pattern 2: (a − b)² = a² − 2ab + b²
  (These match how we just used identities.)

Do they hold for:

• Counting numbers (positive integers)? Yes.
• Negative integers? Yes.
• Fractions (rationals)? Yes.
• In fact, they hold for all real numbers (and even complex numbers), because they are algebraic identities derived from distributive, associative, and commutative properties of addition and multiplication, which are valid across these number systems.

Justification:
Expand using distribution:
(a + b)² = (a + b)(a + b) = a·a + a·b + b·a + b·b = a² + ab + ba + b² = a² + 2ab + b² (since ab = ba).
Similarly,
(a − b)² = (a − b)(a − b) = a² − ab − ba + b² = a² − 2ab + b².
No step assumes a, b are positive or integers; it only uses basic arithmetic laws valid for reals (and rationals, integers, etc.). Therefore, the patterns hold for negative integers and fractions as well.

If by “Pattern 2” you meant the difference-of-squares factorization a² − b² = (a − b)(a + b), that identity also holds for all real numbers (and beyond), by direct expansion of the right-hand side.

Summary:

• 1. Equal: (a − b)² = (b − a)².
• 2. 100 = 26² − 24² = 10² − 0² (and other factor-based pairs).
• 3. 406² = 164,836; 72² = 5,184; 145² = 21,025; 1097² = 1,203,409; 124² = 15,376.
• 4. The standard square identities (and difference of squares) hold for counting numbers, negative integers, and fractions—indeed for all real numbers.