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Directions (32-36) : In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.
Give answer :
(1) \( x \le y \)
(2) \( x \ge y \)
(3) Relationship between X and y cannot be established
(4) x < y
(5) x > y
I.\( 2x^{2} + 15x +27 = 0 \)
II. \( 2y^{2} + 7y +6 = 0 \)
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A) \( x \le y \) |
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B) \( x \ge y \) |
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C) Relationship between X and y cannot be established |
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D) x < y |
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E) x > y |
Correct Answer:
| D) x < y |
Description for Correct answer:
I. \( \large 2x^{2} + 15x + 27 = 0 \)
\( \large 2x^{2} + 9x + 6x + 27 = 0 \)
\( \largex( 2x + 9) + 3(2x + 9)= 0 \)
\( \large (x + 3)(2x + 9) = 0 \)
\( \large x = -\frac{9}{2},-3 \)
II. \( \large 2y^{2} + 7y + 6 = 0 \)
\( \large 2y^{2} + 4y + 3y + 6 = 0 \)
\( \large 2y( y + 2) + 3(y + 2) = 0 \)
\( \large (y + 2)( 2y + 3) = 0 \)
\( \large y = -2, - \frac{3}{2} \)
So x < y
I. \( \large 2x^{2} + 15x + 27 = 0 \)
\( \large 2x^{2} + 9x + 6x + 27 = 0 \)
\( \largex( 2x + 9) + 3(2x + 9)= 0 \)
\( \large (x + 3)(2x + 9) = 0 \)
\( \large x = -\frac{9}{2},-3 \)
II. \( \large 2y^{2} + 7y + 6 = 0 \)
\( \large 2y^{2} + 4y + 3y + 6 = 0 \)
\( \large 2y( y + 2) + 3(y + 2) = 0 \)
\( \large (y + 2)( 2y + 3) = 0 \)
\( \large y = -2, - \frac{3}{2} \)
So x < y
Part of solved IBPS PO-Prelims questions and answers : Exams >> Bank Exams >> IBPS PO-Prelims
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