In the following question two equations numbered I and II are given. You have to solve both the equations and __________ Give answer
I. \( \Large 3x^{2} \) + 8x + 4 = 0
II. \( \Large 4y^{2} \) - 19y + 12 = 0
Correct Answer: Description for Correct answer:
I. \( \Large 3x^{2} \) + 8x + 4 = 0
=> \( \Large 3x^{2} \) + 6x + 2x + 4 = 0
=> 3x ( x + 2 ) + 2 ( x + 2 ) = 0
=> (x + 2 ) ( 3x + 2 ) = 0
\( \Large \therefore \) x = -2 or \( \Large -\frac{2}{3} \)
II. \( \Large 4y^{2} \) - 19y + 12 = 0
=> \( \Large 4y^{2} \) - 16y - 3y + 12 = 0
=> 4y ( y - 4 ) - 3 ( y - 4 ) = 0
=> ( y - 4 ) ( 4y - 3 ) = 0
\( \Large \therefore \) y = 4 or \( \Large \frac{3}{4} \)
Clearly, x < y
Part of solved Linear Equations questions and answers :
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