Question 1

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 2

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 3

If a - b = 7 and ab = 12, find a² + b².

Accepted Answer

Step 1: Use the identity (a - b)² = a² - 2ab + b². Step 2: Substitute a - b = 7 and ab = 12: (7)² = a² - 2(12) + b². Step 3: Simplify: 49 = a² - 24 + b². Step 4: Therefore, a² + b² = 49 + 24 = 73.

Question 4

Expand: (2x + 3y)(2x - 3y)

Accepted Answer

Step 1: Use the difference of squares identity (a + b)(a - b) = a² - b². Step 2: Here, a = 2x and b = 3y. Step 3: Apply: (2x)² - (3y)² = 4x² - 9y².

Question 5

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

Accepted Answer

Step 1: Use the identity (a + b)² = a² + 2ab + b² = 34 + 2(15) = 64. Step 2: Use the identity (a - b)² = a² - 2ab + b² = 34 - 2(15) = 4. Step 3: Add them: 64 + 4 = 68.

Question 6

If x + y = 12 and x² + y² = 90, then find the value of x³ + y³.

Accepted Answer

Step 1: From x + y = 12, square both sides: (x + y)² = 144, so x² + 2xy + y² = 144. Step 2: Since x² + y² = 90, we have 90 + 2xy = 144, giving xy = 27. Step 3: Use the identity x³ + y³ = (x + y)³ - 3xy(x + y) = 12³ - 3(27)(12) = 1728 - 972 = 756.

RRB NTPC Algebra Questions

RRB NTPC Algebra — Past Exam Questions

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