| 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. \[{{x}^{2}}+5x+6=0\] |
| II. \[4{{y}^{2}}+24y+35=0\] |
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: E
Solution :
| I. \[{{x}^{2}}+5x+6=0\] |
| \[\Rightarrow \]\[{{x}^{2}}+3x+2x+6=0\] |
| \[\Rightarrow \]\[x\,\,(x+3)+2\,\,(x+3)=0\] |
| \[\Rightarrow \]\[(x+2)(x+3)=0\] |
| \[\therefore \]\[x=-\,\,2,\]\[-\,\,3\] |
| II. \[4{{y}^{2}}+24y+35=0\] |
| \[\Rightarrow \]\[4{{y}^{2}}+14y+10y+35=0\] |
| \[\Rightarrow \]\[2y\,\,(2y+7)+5\,\,(2y+7)=0\] |
| \[\Rightarrow \] \[(2y+5)(2y+7)=0\] |
| \[\therefore \] \[y=-\frac{5}{2},\]\[-\frac{7}{2}\] |
| Hence, relationship between x and y cannot be established. |
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