Topics
\( \Large 3x^{2}+8x+4=0; 4y^{2}-19y+12=0 \)
|
A) If x>y |
|
B) \( \Large If \ x\ge y \) |
|
C) lfx |
|
D) \( \Large If \ x \le y \) |
Correct Answer:
| D) \( \Large If \ x \le y \) |
Description for Correct answer:
\( \Large 3x^{2} + 8x +4 = 0 \)
= \( \Large 3x^{2}+6x+2x+4=0 \)
= \( \Large 3x \left(x+2\right) + 2 \left(x+2\right) = 0 \)
= \( \Large \left(x+2\right) \left(3x+2\right) = 0 \)
Therefore, x = -2, \( \Large -\frac{2}{3} \)
and \( \Large 4y^{2} - 19y + 12 = 0 \)
=> \( \Large 4y^{2} - 16y -3y + 12 = 0 \)
=> \( \Large 4y \left(y - 4\right) - 3 \left(y - 4\right) = 0 \)
=> \( \Large \left(y - 4\right) \left(4y - 3\right) = 0 \)
Therefore, \( \Large y = 4, \frac{3}{4} \)
Hence, y > x or x < y
Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations
Comments
Similar Questions
