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In the given questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.

I: 3x2 – 15x + 12 = 0

II : 4y2 – 80y + 256 = 0


1. x > y
2. x ≤ y
3. No relation in x and y or x = y
4. x ≥ y
5. x < y

1 Answer

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Best answer
Correct Answer - Option 2 : x ≤ y

Calculation:

I: 3x2 – 15x + 12 = 0

⇒ 3x2 – 12x – 3x + 12 = 0

⇒ 3x(x – 4) – 3(x – 4) = 0

⇒ (x – 4) × (3x – 3) = 0

⇒ x = 4, 1

II: 4y2 – 80y + 256 = 0

⇒ 4y2 – 16y – 64y + 256 = 0

⇒ 4y × (y – 4) – 64 × (y – 4) = 0

⇒ (y – 4) × (4y – 64) = 0

⇒ y = 4, 16

Comparing values of x and y,

Value of x

relation

Value of y

4

=

4

4

16

1

4

1

16

From the table, we can say, x ≤  y.

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