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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