Exploring Algebraic Identities (Class 9): Notes और MCQ Practice

आसान भाषा में

Algebraic Identity एक ऐसा mathematical rule या equation होता है जो variables की हर एक value के लिए हमेशा true रहता है। यह ordinary equation से अलग है क्योंकि equation केवल कुछ खास values के लिए सही होता है। Identities की मदद से हम बड़ी calculations, factorisation और simplification बेहद आसानी से कर सकते हैं।

Key Formulas & Points

• Identity vs Equation: Equation कुछ चुनिंदा values के लिए true होता है (जैसे x² − 1 = 24), जबकि Identity हर value के लिए valid होती है।
• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• a² − b² = (a + b)(a − b) (Śhrīdharāchārya formula: a² = (a + b)(a − b) + b²)
• (x + a)(x + b) = x² + (a + b)x + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a − b)³ = a³ − 3a²b + 3ab² − b³
• x³ − y³ = (x − y)(x² + xy + y²)
• x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

2 Memory Tricks

1. (a − b)³ का Alternating Sign Trick: इसके चार terms हमेशा alternate signs (+ − + −) फॉलो करते हैं: +a³, −3a²b, +3ab², −b³।
2. Middle Term Splitting (SP Rule): x² + (a + b)x + ab में middle term हमेशा factors का Sum (a + b) होता है और constant term उनका Product (ab) होता है।

2 Common Board-Exam Mistakes

1. (a + b)² को a² + b² लिखना: हमेशा याद रखें कि (a + b)² ≠ a² + b²; बीच का 2ab term लिखना कभी न भूलें।
2. (a − b)² में b² का गलत sign: कई बच्चे गलती से a² − 2ab − b² लिख देते हैं। (−b)² हमेशा positive (+b²) ही रहता है!

Exploring Algebraic Identities

1. Chapter in Short

This chapter introduces algebraic identities as universal mathematical rules that remain true for any values of the variables. By moving beyond simple equations, students learn how to:

• Visualize identities geometrically using 2D areas (squares and rectangles) and 3D volumes (cubes and cuboids).
• Expand products of binomials and trinomials efficiently without long multiplication.
• Factorise algebraic expressions by recognizing patterns or splitting terms.
• Perform fast mental math and numerical computations using identities.

2. Key Concepts Explained Simply

Equations vs. Algebraic Identities

• Equation: A mathematical statement that is true only for specific values of the variable.
• Example: x² − 1 = 24 is only true when x = 5 or x = −5.
• Algebraic Identity: An equality that holds true for all possible values of the variables involved.
• Example: (x + y)² = x² + 2xy + y² is true whether x and y are positive, negative, or rational numbers.

Geometric Visualisation of Identities

• Squaring a binomial, (a + b)²: Imagine a square of side length (a + b). Its total area is (a + b)². When divided, it forms:
• One square of area a²
• One square of area b²
• Two rectangles each of area a × b (total area = 2ab)
• Combining these gives: (a + b)² = a² + 2ab + b².
• Cubing a binomial, (a + b)³: Imagine a large cube of side (a + b). Its total volume is (a + b)³. When split apart, it breaks into:
• Two smaller cubes of volumes a³ and b³
• Three cuboids of dimensions a × a × b (volume = 3a²b)
• Three cuboids of dimensions a × b × b (volume = 3ab²)
• Combining these gives: (a + b)³ = a³ + 3a²b + 3ab² + b³.

Factorisation by Splitting the Middle Term

When factorising a quadratic expression like x² + (a + b)x + ab into (x + a)(x + b):

• Look at the middle coefficient: it must be the sum of two numbers (a + b).
• Look at the constant term: it must be the product of those same two numbers (a × b).
• Example: For x² + 7x + 12, we need two numbers that add up to 7 and multiply to 12. Those numbers are 3 and 4. Thus, x² + 7x + 12 = (x + 3)(x + 4).

3. Important Definitions, Formulas, Facts, and Dates

Standard Algebraic Identities

1. Square of a sum: (a + b)² = a² + 2ab + b²
2. Square of a difference: (a − b)² = a² − 2ab + b²
3. Difference of squares: a² − b² = (a + b)(a − b), or rewritten as a² = (a + b)(a − b) + b²
4. Square of a trinomial: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
5. Product of two linear factors: (x + a)(x + b) = x² + (a + b)x + ab
6. Cube of a sum: (a + b)³ = a³ + 3a²b + 3ab² + b³
7. Cube of a difference: (a − b)³ = a³ − 3a²b + 3ab² − b³ (Note: Signs alternate: +, −, +, −)
8. Sum of cubes factorisation: x³ + y³ = (x + y)(x² − xy + y²)
9. Difference of cubes factorisation: x³ − y³ = (x − y)(x² + xy + y²)
10. Three-variable cubic identity: (x + y + z)(x² + y² + z² − xy − yz − zx) = x³ + y³ + z³ − 3xyz

Historical Facts and Special Patterns

• Historical Method (750 CE): In 750 CE, the Indian mathematician Śhrīdharāchārya proposed using the identity a² = (a + b)(a − b) + b² to quickly compute squares of numbers.
• Consecutive Squares Property: If you take three consecutive squares (n − 1)², n², and (n + 1)², adding the smallest and largest and subtracting twice the middle square always equals 2:
• [(n − 1)² + (n + 1)²] − 2n² = (2n² + 2) − 2n² = 2.

4. Solved Examples

Example 1: Numerical Calculation Using Śhrīdharāchārya's Method

Question: Compute 55² using the identity a² = (a + b)(a − b) + b².

• Solution:
• Let a = 55 and choose b = 5.
• 55² = (55 + 5)(55 − 5) + 5²
• 55² = (60 × 50) + 25
• 55² = 3000 + 25 = 3025.

Example 2: Factoring with a Common Factor First

Question: Factorise 50p² + 60pq + 18q² completely.

• Solution:
• Notice that 2 is a common factor in all three terms:

50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²)

• Now compare 25p² + 30pq + 9q² to a² + 2ab + b²:
• a² = (5p)², so a = 5p
• b² = (3q)², so b = 3q
• Check 2ab: 2 × (5p) × (3q) = 30pq (matches)
• Thus, the factored form is 2(5p + 3q)².

Example 3: Factoring Quadratic Expressions Without Tiles

Question: Factorise x² − 5x + 6.

• Solution:
• Compare with x² + (a + b)x + ab:
• a + b = −5
• ab = 6
• Look for factors of 6 that add up to −5: these are −2 and −3.
• Therefore, x² − 5x + 6 = (x − 2)(x − 3).

Example 4: Identifying the Side of a Cube from Its Volume

Question: Find the edge length of a cube whose volume is p³ + 6p²q + 12pq² + 8q³.

• Solution:
• Compare with the expansion of (a + b)³ = a³ + 3a²b + 3ab² + b³:
• a³ = (p)³ → a = p
• b³ = 8q³ = (2q)³ → b = 2q
• Check middle terms:
• 3a²b = 3(p)²(2q) = 6p²q
• 3ab² = 3(p)(2q)² = 12pq²
• The expression equals (p + 2q)³.
• Therefore, the side length of the cube is (p + 2q) units.

Example 5: Application of Cubic Identities

Question: The sum of three numbers is 10, their product is 25, and the sum of their squares is 38. Find the sum of their cubes.

• Solution:
• Given: x + y + z = 10, xyz = 25, and x² + y² + z² = 38.
• Using the identity (x + y + z)² = x² + y² + z² + 2(xy + yz + zx):
• 10² = 38 + 2(xy + yz + zx)
• 100 − 38 = 2(xy + yz + zx)
• 62 = 2(xy + yz + zx) → xy + yz + zx = 31.
• Now use the cubic identity: x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − (xy + yz + zx))
• x³ + y³ + z³ − 3(25) = 10 × (38 − 31)
• x³ + y³ + z³ − 75 = 10 × 7
• x³ + y³ + z³ − 75 = 70
• x³ + y³ + z³ = 70 + 75 = 145.

5. Common Mistakes and Quick Revision Points

Common Mistakes to Avoid

• Forgetting the middle term: Never write (a + b)² = a² + b². Always remember to include the middle term 2ab.
• Sign errors in (a − b)³: Students often make mistakes with the signs. Remember that the signs alternate:
• Positive first term: +a³
• Negative second term: −3a²b
• Positive third term: +3ab²
• Negative fourth term: −b³
• Skipping the greatest common factor: In problems like 50p² + 60pq + 18q², do not try to find square roots of 50 directly. Always check if a common number can be factored out first (here, factor out 2).
• Confusing sum of cubes with cube of a sum:
• (a + b)³ is NOT the same as a³ + b³.
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• a³ + b³ = (a + b)(a² − ab + b²)

Quick Revision Points

• Identities work for all numbers: positive, negative, and rational.
• When factorising x² + Bx + C, find two numbers whose product is C and whose sum is B.
• The identity (a + b + c)² always has 6 terms: 3 squared terms (a² + b² + c²) and 3 double-product terms (2ab + 2bc + 2ca).