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In the following questions two equations I and II are given. You have to solve both the equations and Give answer
I. \( \Large 6x^{2} \) + 13x + 5 = 0
II. \( \Large 9y^{2} \) + 22y + 8 = 0
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A) x > y |
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B) x \( \Large \leq \) y |
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C) x < y |
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D) x = y or relationship between x and y cannot be established |
Correct Answer:
| D) x = y or relationship between x and y cannot be established |
Description for Correct answer:
I. 6 \( \Large x^{2} \) + 13x + 5 = 0
=> 6 \( \Large x^{2} \) + 10x + 3x + 5 = 0
=> 2x ( 3x + 5 ) + 1 ( 3x + 5 ) = 0
=> ( 3x + 5 ) ( 2x + 1 ) = 0
=> x = \( \Large \frac{-5}{3} \) or \( \Large - \frac{1}{2} \)
II. 9 \( \Large y^{2} \) + 22y + 8 = 0
=> 9 \( \Large y^{2} \) + 18y + 4y + 8 = 0
=> 9y ( y + 2 ) + 4 ( y + 2 ) = 0
=> ( y + 2 ) ( 9y + 4 ) = 0
=> y = -2 or \( \Large \frac{-4}{9} \)
I. 6 \( \Large x^{2} \) + 13x + 5 = 0
=> 6 \( \Large x^{2} \) + 10x + 3x + 5 = 0
=> 2x ( 3x + 5 ) + 1 ( 3x + 5 ) = 0
=> ( 3x + 5 ) ( 2x + 1 ) = 0
=> x = \( \Large \frac{-5}{3} \) or \( \Large - \frac{1}{2} \)
II. 9 \( \Large y^{2} \) + 22y + 8 = 0
=> 9 \( \Large y^{2} \) + 18y + 4y + 8 = 0
=> 9y ( y + 2 ) + 4 ( y + 2 ) = 0
=> ( y + 2 ) ( 9y + 4 ) = 0
=> y = -2 or \( \Large \frac{-4}{9} \)
Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations
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