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Direction: In the following question two equation numbered I and II are given. Solve the equation and answer the question: (i) 3x2 - 22x + 24 = 0 (ii) 3y2 - 20y + 25 = 0

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Direction: In the following question two equation numbered I and II are given. Solve the equation and answer the question:

(i) 3x2 - 22x + 24 = 0

(ii) 3y2 - 20y + 25 = 0
1). x > y
2). x > y
3). x < y
4). x < y


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Answered by on | Votes 0 |

i. 3x2 - 22x + 24 = 0

⇒ 3x2 - 18x - 4x + 24 = 0

⇒ 3x (x - 6) - 4(x - 6) = 0

⇒ (x - 6) (3x - 4) = 0

⇒ x = 6 or 4/3

ii. 3y2 - 20y + 25 = 0

⇒ 3y2 - 15y - 5y + 25 = 0

⇒ 3y (y - 5) - 5(y - 5) = 0

⇒ (y - 5) (3y - 5) = 0

⇒ y = 5 or 5/3

When, x = 6, x > y for y = 5 and x > y for y = 5/3

And when x = 4/3, x < y for y = 5 and x < y for y = 5/3

∴ the relationship cannot be established between x and y

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