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vishnupreethi

· started a discussion

· 1 Months ago

Solution provided is not convincing (Question: 111646.0 )

when-2 is greater than -3, it explicitly states that x is lesser than or equal to y

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 + 7x + 12 = 0

II. y2 + 6y + 8 = 0

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

From equation I,

x2 + 7x + 12 = 0

x2 + 3x + 4x + 12 = 0

(x + 3) (x + 4) = 0

x = -3,  -4

From equation II,

y2 + 6y + 8 = 0

y2 + 2y + 4y + 8=0

⇒ (y +2) (y + 4) = 0

y = -2, -4

∴, No relation

akash

· commented

· 1 Months ago

b is answer

kanhaiya jha

· commented

· 1 Months ago

From equation I,

x^2 + 7x + 12 = 0

x^2 + 3x + 4x + 12 = 0

(x + 3) (x + 4) = 0

x = -3, -4

From equation II,

y^2 + 6y + 8 = 0

y^2 + 2y + 4y + 8=0

⇒ (y +2) (y + 4) = 0

y = -2, -4

-3<-2 means x<y
-3>-4 means x>y


∴, No relation
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