Question 1

If x + 1/x = 5, find x² + 1/x².

Accepted Answer

Step 1: Use the identity (x + 1/x)² = x² + 2(x)(1/x) + 1/x² = x² + 2 + 1/x². Step 2: Rearrange: x² + 1/x² = (x + 1/x)² - 2. Step 3: Substitute x + 1/x = 5: x² + 1/x² = 5² - 2 = 25 - 2 = 23.

Question 2

If (a + b)² = 81 and ab = 18, find the value of a² + b².

Accepted Answer

Step 1: Use the identity (a + b)² = a² + 2ab + b². Step 2: Substitute (a + b)² = 81 and ab = 18 into the identity: 81 = a² + 2(18) + b². Step 3: Simplify: 81 = a² + 36 + b². Step 4: Therefore, a² + b² = 81 - 36 = 45.

Question 3

If p - q = 7 and pq = 12, find p² + q².

Accepted Answer

Step 1: Use the identity (p - q)² = p² - 2pq + q². Step 2: Rearrange to get p² + q² = (p - q)² + 2pq. Step 3: Substitute (p - q) = 7 and pq = 12: p² + q² = 7² + 2(12) = 49 + 24 = 73.

Question 4

Simplify: (a + b + c)² - (a² + b² + c²)

Accepted Answer

Step 1: Expand (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. Step 2: Subtract (a² + b² + c²): (a² + b² + c² + 2ab + 2bc + 2ca) - (a² + b² + c²) = 2ab + 2bc + 2ca. Step 3: Factor: 2(ab + bc + ca).

Question 5

Simplify: (x + 5)² - (x - 5)²

Accepted Answer

Step 1: Use the identity (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b². Step 2: Expand (x + 5)² = x² + 10x + 25. Step 3: Expand (x - 5)² = x² - 10x + 25. Step 4: Subtract: (x² + 10x + 25) - (x² - 10x + 25) = x² + 10x + 25 - x² + 10x - 25 = 20x.

Question 6

Expand: (2m + 3n)(2m - 3n)

Accepted Answer

Step 1: Use the difference of squares identity (a + b)(a - b) = a² - b². Step 2: Here a = 2m and b = 3n. Step 3: Apply: (2m + 3n)(2m - 3n) = (2m)² - (3n)² = 4m² - 9n².

Question 7

Simplify: (p + q + r)² - (p² + q² + r²) using algebraic identities.

Accepted Answer

Step 1: Expand (p + q + r)² = p² + q² + r² + 2pq + 2qr + 2rp. Step 2: Subtract (p² + q² + r²) from both sides: (p + q + r)² - (p² + q² + r²) = 2pq + 2qr + 2rp. Step 3: Factor: 2(pq + qr + rp).

Question 8

If (a + b)² = 81 and (a - b)² = 25, find the value of ab.

Accepted Answer

Step 1: Expand both identities: (a + b)² = a² + 2ab + b² = 81 and (a - b)² = a² - 2ab + b² = 25. Step 2: Subtract the second from the first: (a² + 2ab + b²) - (a² - 2ab + b²) = 81 - 25, which gives 4ab = 56. Step 3: Therefore, ab = 14.

Question 9

Simplify: (x + 1/x)² - (x - 1/x)² using algebraic identities.

Accepted Answer

Step 1: Use the identity A² - B² = (A + B)(A - B). Here A = (x + 1/x) and B = (x - 1/x). Step 2: A + B = (x + 1/x) + (x - 1/x) = 2x. Step 3: A - B = (x + 1/x) - (x - 1/x) = 2/x. Step 4: Therefore, (x + 1/x)² - (x - 1/x)² = 2x × (2/x) = 4.

Question 10

If a² + b² = 34 and ab = 15, find the value of (a + b)².

Accepted Answer

Step 1: Use the identity (a + b)² = a² + 2ab + b². Step 2: Rearrange: (a + b)² = (a² + b²) + 2ab. Step 3: Substitute the given values: (a + b)² = 34 + 2(15) = 34 + 30 = 64.

Question 11

If x + 1/x = 5, find the value of x² + 1/x².

Accepted Answer

Step 1: Square both sides of x + 1/x = 5: (x + 1/x)² = 25. Step 2: Expand using (a + b)² = a² + 2ab + b²: x² + 2(x)(1/x) + 1/x² = 25. Step 3: Simplify: x² + 2 + 1/x² = 25. Step 4: Therefore, x² + 1/x² = 25 - 2 = 23.

Question 12

If (a + b + c) = 12 and (ab + bc + ca) = 47, find the value of a² + b² + c².

Accepted Answer

Step 1: Use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Step 2: Substitute the given values: 12² = a² + b² + c² + 2(47). Step 3: Simplify: 144 = a² + b² + c² + 94. Step 4: Therefore, a² + b² + c² = 144 - 94 = 50.

Question 13

If a + b + c = 12 and a² + b² + c² = 56, find the value of ab + bc + ca.

Accepted Answer

Step 1: Use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Step 2: Substitute the given values: 12² = 56 + 2(ab + bc + ca). Step 3: Simplify: 144 = 56 + 2(ab + bc + ca), so 2(ab + bc + ca) = 88. Step 4: Therefore ab + bc + ca = 44.

Question 14

If x + 1/x = 5, find the value of x³ + 1/x³.

Accepted Answer

Step 1: Square the given equation: (x + 1/x)² = 25, which expands to x² + 2 + 1/x² = 25, so x² + 1/x² = 23. Step 2: Use the identity x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x). Step 3: Substitute: x³ + 1/x³ = 5³ − 3(5) = 125 − 15 = 110.

Question 15

If (a + b)² = 81 and (a − b)² = 49, find the value of ab.

Accepted Answer

Step 1: Expand both identities: (a + b)² = a² + 2ab + b² = 81 and (a − b)² = a² − 2ab + b² = 49. Step 2: Subtract the second from the first: (a² + 2ab + b²) − (a² − 2ab + b²) = 81 − 49, which gives 4ab = 32. Step 3: Therefore ab = 8.

Question 16

If a² + b² + c² = 50 and ab + bc + ca = 25, find the value of (a + b + c)².

Accepted Answer

Step 1: Use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Step 2: Substitute the given values: (a + b + c)² = 50 + 2(25) = 50 + 50 = 100.

Question 17

If (p + q)³ = 216 and (p − q)³ = 64, find the value of p.

Accepted Answer

Step 1: Take cube roots: p + q = 6 and p − q = 4. Step 2: Add the two equations: (p + q) + (p − q) = 6 + 4, which gives 2p = 10. Step 3: Therefore p = 5.

Question 18

If x² + y² = 34 and xy = 15, find the value of (x + y)² − (x − y)².

Accepted Answer

Step 1: Expand (x + y)² = x² + 2xy + y² and (x − y)² = x² − 2xy + y². Step 2: Compute the difference: (x + y)² − (x − y)² = (x² + 2xy + y²) − (x² − 2xy + y²) = 4xy. Step 3: Substitute xy = 15: (x + y)² − (x − y)² = 4(15) = 60.

IBPS Clerk Algebraic Identities

IBPS Clerk Algebraic Identities — Past Exam Questions

Algebraic Identities— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Algebraic Identities