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avinash

· started a discussion

· 1 Months ago

multiple answers (Question: 107105.0 )

for equation y 6 and -1 also satisfies.

Question:
DIRECTION : In the given question two equations numbered I and II are given. You have to solve both the equations and Give answer if


I. x2 – 14x + 48 = 0

II. y2 + 6 = 5y

Options:
A) X > Y
B) X ≥ Y
C) X < Y
D) X ≤ Y
E) X = Y or the relationship cannot be established
Solution:

I. x2 – 14x + 48 = 0

⇒ x2 – 8x – 6x + 48 = 0

⇒ x (x – 8) -6 (x – 8) = 0

⇒ (x – 6) (x – 8) = 0

∴ X = 6, 8

II. y2 -5y + 6 = 0

⇒ y2 – 3y -2y + 6 = 0

⇒ y(y – 3) -2(y – 3) = 0

⇒ (y – 2) (y – 3) = 0

∴ y = 2, 3

∴ x > y

kanhaiya jha

· commented

· 1 Months ago

Why????..........Take y=6 and -1....put in equation: y2 -5y + 6 = 0......It can not be zero...then how can it satisfy....

See the detail sol:

I. x2 – 14x + 48 = 0
⇒ x2 – 8x – 6x + 48 = 0
⇒ x (x – 8) -6 (x – 8) = 0
⇒ (x – 6) (x – 8) = 0
∴ X = 6, 8
II. y2 -5y + 6 = 0
⇒ y2 – 3y -2y + 6 = 0
⇒ y(y – 3) -2(y – 3) = 0
⇒ (y – 2) (y – 3) = 0
∴ y = 2, 3
∴ x > y
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