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ABHIJEET PATIL

· started a discussion

· 1 Months ago

i think its wrong (Question: 90253.2 )

if x=3 also y=3 then there should be the case x=y, for x=4 there is always x>y but as x=3 is also possibility hence x>=y would b the right ans

Question:
DIRECTIONS: In the following questions two equations numbered I and II are given. You have to solve both the equations and indicate the correct answer.

I.   x2 - 7x + 12 = 0


II. y2 + y - 12 = 0
Options:
A) if x > y
B) if x ≥ y
C) if x < y
D) if x ≤ y
E) if x = y or the relationship cannot be established
Solution:

I. x2 - 4x - 3x + 12 = 0

⇒ x (x - 4) -3 (x - 4) = 0

⇒ (x - 3) (x - 4) = 0

∴ x = 3 or 4

II. y2 + 4y - 3y - 12 = 0

⇒ y (y + 4) - 3(y + 4) = 0

⇒ (y - 3) (y + 4) = 0

∴ y = 3 or – 4

Clearly, x > y

Knowledge Expert

· commented

· 1 Months ago

I. x2 - 4x - 3x + 12 = 0
⇒ x (x - 4) -3 (x - 4) = 0
⇒ (x - 3) (x - 4) = 0
∴ x = 3 or 4
II. y2 + 4y - 3y - 12 = 0
⇒ y (y + 4) - 3(y + 4) = 0
⇒ (y - 3) (y + 4) = 0
∴ y = 3 or – 4
Clearly, x >= y


option(b) is correct....
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