In the following question two equations numbered I and II are given. You have to solve both the equations and Give answer

I. \( \Large x^{2} \) - x - 12 = 0

II. \( \Large y^{2} \) + 5y + 6 = 0

A) x > y

B) x \( \Large \geq \) y

C) x < y

D) x \( \Large \leq \) y

Correct answer:
B) x \( \Large \geq \) y

Description for Correct answer:
I. \( \Large x^{2} \) - x - 12 = 0

=> \( \Large x^{2} \) - 4x + 3x - 12 = 0

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

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

\( \Large \therefore \) x = 4 or -3

II. \( \Large y^{2} \) + 5y + 6 = 0

=> \( \Large y^{2} \) + 3y + 2y + 6 = 0

=> y ( y + 3 ) + 2 ( y + 3 ) = 0

=> ( y + 3 ) ( y + 2 ) = 0

\( \Large \therefore \) y = -3 or -2

Clearly, x \( \Large \geq \) y

Please provide the error details in above question

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