Are you looking for complete, step-by-step verified solutions for DAV Class 8 Maths Chapter 7 Enrichment Questions (Algebraic Identities)? At NCERTBooksPDF, our detailed solutions provide clear proofs and step-by-step methods for reciprocal identity problems, sum-product systems, and advanced identity applications.
Key Identities Used in Enrichment Problems
| Identity Form | Formula |
|---|---|
| Identity I | (a + b)2 = a2 + 2ab + b2 |
| Identity II | (a – b)2 = a2 – 2ab + b2 |
| Reciprocal Sum | (x + 1/x)2 = x2 + 1/x2 + 2 |
| Reciprocal Difference | (x – 1/x)2 = x2 + 1/x2 – 2 |
Chapter 7 Enrichment Questions: Complete Solutions
Question 1: Reciprocal Powers
Problem: If x + 1/x = 5, find the value of:
(i) x2 + 1/x2
(ii) x4 + 1/x4
Solution:
(i) Finding x2 + 1/x2:
Squaring both sides of x + 1/x = 5:
(x + 1/x)2 = 52
x2 + 2(x)(1/x) + (1/x)2 = 25
x2 + 2 + 1/x2 = 25
x2 + 1/x2 = 25 – 2 = 23.
(ii) Finding x4 + 1/x4:
Squaring both sides of x2 + 1/x2 = 23:
(x2 + 1/x2)2 = 232
(x2)2 + 2(x2)(1/x2) + (1/x2)2 = 529
x4 + 2 + 1/x4 = 529
x4 + 1/x4 = 529 – 2 = 527.
Question 2: Difference of Reciprocals
Problem: If x – 1/x = 7, find the value of x2 + 1/x2.
Solution:
Squaring both sides of x – 1/x = 7:
(x – 1/x)2 = 72
x2 – 2(x)(1/x) + 1/x2 = 49
x2 – 2 + 1/x2 = 49
x2 + 1/x2 = 49 + 2 = 51.
Question 3: Linear Combination System
Problem: If 2x + 3y = 12 and xy = 6, find the value of 4x2 + 9y2.
Solution:
Squaring both sides of 2x + 3y = 12:
(2x + 3y)2 = 122
(2x)2 + 2(2x)(3y) + (3y)2 = 144
4x2 + 12xy + 9y2 = 144
Substitute xy = 6:
4x2 + 12(6) + 9y2 = 144
4x2 + 72 + 9y2 = 144
4x2 + 9y2 = 144 – 72 = 72.
Question 4: Symmetric Values
Problem: If a2 + b2 = 29 and ab = 10, find the values of (i) a + b and (ii) a – b.
Solution:
(i) Finding a + b:
(a + b)2 = a2 + b2 + 2ab
= 29 + 2(10) = 29 + 20 = 49
a + b = √49 = ±7 (or 7 for positive values).
(ii) Finding a – b:
(a – b)2 = a2 + b2 – 2ab
= 29 – 2(10) = 29 – 20 = 9
a – b = √9 = ±3 (or 3 for a > b).
Question 5: Continuous Identity Multiplication
Problem: Show that: (x + 1)(x – 1)(x2 + 1)(x4 + 1) = x8 – 1.
Proof:
Step 1: (x + 1)(x – 1) = x2 – 1.
Step 2: (x2 – 1)(x2 + 1) = (x2)2 – 12 = x4 – 1.
Step 3: (x4 – 1)(x4 + 1) = (x4)2 – 12 = x8 – 1.
Hence Proved!
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