| Directions: In each of these questions two equations I and II are given. You have to solve both the equations and give answer. [IBPS (SO) 2014] |
| I. \[2{{x}^{2}}+5x+3=0\] |
| II. \[{{y}^{2}}+9y+14=0\] |
A) If \[x\ge y\]
B) If \[x>y\]
C) If \[x\le y\]
D) If \[x<y\]
E) If relationship between x and y cannot be established
Correct Answer: B
Solution :
| I. \[2{{x}^{2}}+5x+3=0\] |
| \[\Rightarrow \]\[2{{x}^{2}}+2x+3x+3=0\] |
| \[\Rightarrow \]\[2x\,\,(x+1)+3\,\,(x+1)=0\] |
| \[\Rightarrow \]\[(x+1)(2x+3)=0\] |
| \[\therefore \]\[x=-\frac{3}{2},\]\[-\,\,1\] |
| II. \[{{y}^{2}}+9y+14=0\] |
| \[\Rightarrow \]\[{{y}^{2}}+7y+2y+14=0\] |
| \[\Rightarrow \]\[y\,\,(y+7)+2\,\,(y+7)=0\] |
| \[\Rightarrow \]\[(y+2)(y+7)=0\] |
| \[\therefore \]\[y=-\,\,2,\]\[-\,\,7\] |
| Hence, \[x>y\] |
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