| Directions: In these questions two equations numbered I and II are given. |
| You have to solve both the equations and give answer. |
| I. \[24{{x}^{2}}+38x+15=0\] |
| II. \[12{{y}^{2}}+28y+15=0\] |
A) If \[x\le y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x>y\]
E) If relationship between x and .y cannot be established
Correct Answer: B
Solution :
| I. \[24{{x}^{2}}+38x+15=0\] |
| \[\Rightarrow \]\[24{{x}^{2}}+20x+18x+15=0\] |
| \[\Rightarrow \]\[4x\,\,(6x+5)+3\,\,(6x+5)=0\] |
| \[\Rightarrow \] \[(4x+3)(6x+5)=0\] |
| \[\therefore \] \[x=-\frac{3}{4},\]\[-\frac{5}{6}\] |
| II. \[12{{y}^{2}}+28y+15=0\] |
| \[\Rightarrow \]\[12{{y}^{2}}+18y+10y+15=0\] |
| \[\Rightarrow \]\[6y\,\,(2y+3)+5\,\,(2y+3)=0\] |
| \[\Rightarrow \] \[(6y+5)(2y+3)=0\] |
| \[\therefore \] \[y=-\frac{3}{2},\]\[-\frac{5}{6}\] |
| Hence, \[x\ge y\] |
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