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In the following question two equations are given. You have to solve the equations and Give answer
I. \( \Large 4x^{2} \) - 8x + 3 = 0
II. \( \Large 2y^{2} \) - 7y + 6 = 0
A) x < y
B) x \( \Large \leq \) y
C) x = y
D) x \( \Large \geq \) y
Correct answer:
B) x \( \Large \leq \) y
Description for Correct answer:
I. 4 \( \Large x^{2} \) - 8x + 3 =0
=> 4 \( \Large x^{2} \) - 6x - 2x + 3 = 0
=> 2x ( 2x - 3 ) - 1 ( 2x - 3 ) = 0
=> ( 2x - 1 ) ( 2x - 3 ) = 0
\( \Large \therefore \) x = \( \Large \frac{1}{2} \) or \( \Large \frac{3}{2} \)
II. 2 \( \Large y^{2} \) - 7y + 6 = 0
=> 2 \( \Large y^{2} \) - 4y - 3y + 6 = 0
=> 2y ( y - 2 ) - 3 ( y - 2 ) = 0
=> ( 2y - 3 ) ( y - 2 ) = 0
\( \Large \therefore \) y = 2 or \( \Large \frac{3}{2} \)
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