| Directions: In these questions two equations numbered I and II are given. You have to solve both the equations and give answer. [IBPS (SO) 2014] |
| I. \[6{{x}^{2}}+5x+1=0\] |
| II. \[15{{y}^{2}}+8y+1=0\] |
A) If \[x>y\]
B) If \[x\le y\]
C) If \[x<y\]
D) If \[x\ge y\]
E) If relationship between x and y cannot be established
Correct Answer: B
Solution :
| I. \[6{{x}^{2}}+5x+1=0\] |
| \[\Rightarrow \]\[6{{x}^{2}}+3x+2x+1=0\] |
| \[\Rightarrow \]\[3x\,\,(2x+1)+1\,\,(2x+1)=0\] |
| \[\Rightarrow \]\[(3x+1)+(2x+1)=0\] |
| \[x=\frac{-1}{3},\]\[-\frac{1}{2}\] |
| II. \[15{{y}^{2}}+8y+1=0\] |
| \[\Rightarrow \]\[15{{y}^{2}}+5y+3y+1=0\] |
| \[\Rightarrow \]\[5y\,\,(3y+1)+1\,\,(3y+1)=0\] |
| \[\Rightarrow \] \[(5y+1)(3y+1)=0\] |
| \[\therefore \] \[y=-\frac{1}{5},\]\[-\frac{1}{3}\] |
| Hence, \[x\le y\] |
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