Question 8 of 27intermediate⚖️ EvaluateLong Answer4 marks

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

Correct Answer

Pattern 1: 2(a2+b2)=(a+b)2+(a−b)22(a^2 + b^2) = (a + b)^2 + (a - b)^2

Pattern 2: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)

Answer: Both patterns hold for all integers (positive and negative) and for fractions as well.

Justification: These are algebraic identities derived from the distributive property, which holds for all real numbers.

Verification with Negative Integers: Let a=−3a = -3, b=2b = 2

Pattern 1: LHS: 2((−3)2+22)=2(9+4)=262((-3)^2 + 2^2) = 2(9 + 4) = 26 RHS: (−3+2)2+(−3−2)2=(−1)2+(−5)2=1+25=26(-3 + 2)^2 + (-3 - 2)^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 LHS = RHS ✓

Pattern 2: LHS: (−3)2−22=9−4=5(-3)^2 - 2^2 = 9 - 4 = 5 RHS: (−3+2)(−3−2)=(−1)(−5)=5(-3 + 2)(-3 - 2) = (-1)(-5) = 5 LHS = RHS ✓

Verification with Fractions: Let a=12a = \frac{1}{2}, b=13b = \frac{1}{3}

Pattern 1: LHS: 2(14+19)=13182\left(\frac{1}{4} + \frac{1}{9}\right) = \frac{13}{18} RHS: (56)2+(16)2=2536+136=2636=1318\left(\frac{5}{6}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{25}{36} + \frac{1}{36} = \frac{26}{36} = \frac{13}{18} LHS = RHS ✓

Exercise: Figure it Out (Section 2) | Q: 4 | (Page Page 14)
For More Understanding

Explanation

The chapter establishes that algebraic identities derived from the distributive property hold for all real numbers, including negative integers and fractions. The identities are proven algebraically and do not depend on the specific values substituted.

Solution Steps

  1. Pattern 1: 2(a2+b2)=(a+b)2+(a−b)22(a^2 + b^2) = (a + b)^2 + (a - b)^2

  2. LHS: 2a2+2b22a^2 + 2b^2

  3. RHS: a2+2ab+b2+a2−2ab+b2=2a2+2b2a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 2a^2 + 2b^2

  4. LHS = RHS for all values of a and b

  5. Pattern 2: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)

  6. RHS: a2−ab+ba−b2=a2−b2a^2 - ab + ba - b^2 = a^2 - b^2

  7. Both are identities, valid for all real numbers

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