| Directions: In each of these questions, two equations I and II are given. You have to solve both the equations and give answer. [IBPS (PO) 2013] |
| I. \[2{{x}^{2}}+3x-20=0\] |
| II. \[2{{y}^{2}}+19y+44=0\] |
A) If \[x<y\]
B) If \[x>y\]
C) If \[x=y\]
D) If \[x\ge y\]
E) If \[x\le y\]or no relationship can be established between x and y
Correct Answer: D
Solution :
| I. \[2{{x}^{2}}+3x-20=0\] |
| \[\Rightarrow \]\[2{{x}^{2}}+8x-5x-20=0\] |
| \[\Rightarrow \]\[2x\,\,(x+4)-5\,\,(x+4)=0\] |
| \[\Rightarrow \] \[(2x-5)(x+4)=0\] |
| \[\therefore \] \[x=\frac{5}{2},\]\[-\,\,4\] |
| II. \[2{{y}^{2}}+19y+44=0\] |
| \[\Rightarrow \]\[2{{y}^{2}}+11y+8y+44=0\] |
| \[\Rightarrow \]\[y\,\,(2y+11)+4\,\,(2y+11)=0\] |
| \[\Rightarrow \]\[(y+4)(2y+11)=0\] |
| \[\therefore \] \[y=-\,\,4,\]\[-\frac{11}{2}\] |
| Hence, \[x\ge y\] |
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