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shraddha

· started a discussion

· 1 Months ago

quadratic equation doubt (Question:100166.0)

I. x2 – x – 12 = 0

⇒ x2 – 4x + 3x – 12 = 0

⇒ x (x – 4)+ 3(x – 4) = 0

⇒ (x – 4) (x + 3) = 0

∴ x = 4 or – 3

II. y2 + 5y + 6 = 0

⇒ y2 + 3y + 2y + 6 = 0

⇒ y (y+3) + 2 (y+3) = 0

⇒ (y + 3) (y+2) =0

∴ y = – 3 or – 2

the answer for this should be x>=y but the answer says x=y relationship cannot be determined. how is that??

Question:
DIRECTIONS: 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 – x – 12 = 0


II. y2 + 5y + 6 = 0 

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 – x – 12 = 0

⇒ x2 – 4x + 3x – 12 = 0

⇒ x (x – 4)+ 3(x – 4) = 0

⇒ (x – 4) (x + 3) = 0

∴ x = 4 or – 3

II. y2 + 5y + 6 = 0

⇒ y2 + 3y + 2y + 6 = 0

⇒ y (y+3) + 2 (y+3) = 0

⇒ (y + 3) (y+2) =0

∴ y = – 3 or – 2 

Knowledge Expert

· commented

· 1 Months ago

Dear student
Given answer is correct
According to question,
x = 4 or -3 y = -3 or -2
at compare,
x y
4 > -3
4 > -2
-3 = -3
-3 < -2
so, Relationship cannot be established.

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