In the given question, two equations numbered l and II are given. You

1.	In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-I. 2x2 - 14x + 20 = 0II. 8y2 - 32y + 24 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relation cannot be determined
Answer» Correct Answer - Option 5 : x = y or the relation cannot be determined Given: I. 2x² – 14x + 20 = 0 II. 8y² – 32y + 24 = 0 Concept used: Using factorization (Middle term split) method. Calculation: 2x² – 14x + 20 = 0 ⇒ 2x² – 10x – 4x + 20 = 0 ⇒ 2x(x – 5) – 4(x – 5) = 0 ⇒ (x – 5)(2x – 4) = 0 ⇒ (x – 5) = 0 or (2x – 4) = 0 ⇒ x = 5 or x = 2 ⇒ x = 2, 5 8y² – 32y + 24 = 0 ⇒ 8y² – 24y – 8y + 24 = 0 ⇒ 8y(y – 3) – 8(y – 3) = 0 ⇒ (y – 3)(8y – 8) = 0 ⇒ (y – 3) = 0 or (8y – 8) = 0 ⇒ y = 3 or y = 1 ⇒ y = 1, 3 ∴ The relation cannot be determined.

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-I. 2x2 - 14x + 20 = 0II. 8y2 - 32y + 24 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relation cannot be determined

Answer» Correct Answer - Option 5 : x = y or the relation cannot be determined

Given:

I. 2x² – 14x + 20 = 0

II. 8y² – 32y + 24 = 0

Concept used:

Using factorization (Middle term split) method.

Calculation:

2x² – 14x + 20 = 0

⇒ 2x² – 10x – 4x + 20 = 0

⇒ 2x(x – 5) – 4(x – 5) = 0

⇒ (x – 5)(2x – 4) = 0

⇒ (x – 5) = 0 or (2x – 4) = 0

⇒ x = 5 or x = 2

⇒ x = 2, 5

8y² – 32y + 24 = 0

⇒ 8y² – 24y – 8y + 24 = 0

⇒ 8y(y – 3) – 8(y – 3) = 0

⇒ (y – 3)(8y – 8) = 0

⇒ (y – 3) = 0 or (8y – 8) = 0

⇒ y = 3 or y = 1

⇒ y = 1, 3

∴ The relation cannot be determined.

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