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In the following question two equations numbered I and II are given. You have to solve both the equations and Give answer
I.\( \Large x^{2} \) - 4 = 0
II.\( \Large y^{2} \) + 6y + 9 = 0
A) x > y
B) x \( \Large \geq \) y
C) x < y
D) x \( \Large \leq \) y
Correct answer:
A) x > y
Description for Correct answer:
I. ( x - 2 )( x + 2 ) = 0
\( \Large \therefore \) x = 2 or -2
II. \( \Large y^{2} \) + 2.y.3 + 9 = 0
=> \( \Large ( y + 3 )^{2} = 0 \)
=> y + 3 = 0
\( \Large \therefore \) y = -3
Clearly, x > y
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