| Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. [IBPS (Officer Recruitment) 2015] |
| Give answer |
| I. \[5{{x}^{2}}+29x+20=0\] |
| II. \[25{{y}^{2}}+25y+6=0\] |
A) if \[x\ge y\]
B) if \[x\le y\]
C) if \[x<y\]
D) if \[x>y\]
E) if relationship between x and y cannot be established
Correct Answer: C
Solution :
| I. \[5{{x}^{2}}+29x+20=0\] |
| \[\Rightarrow \] \[5{{x}^{2}}+25x+4x+20=0\] |
| \[\Rightarrow \] \[5x\,(x+5)+4\,(x+5)=0\] |
| \[\Rightarrow \] \[(5x+4)(x+5)=0\]\[\Rightarrow \]\[x=-\,5\] or \[\frac{-4}{5}\] |
| II. \[25{{y}^{2}}+25y+6=0\] |
| \[\Rightarrow \] \[25{{y}^{2}}+15y+10y+6=0\] |
| \[\Rightarrow \] \[5y\,(5y+3)+2\,(5y+3)=0\] |
| \[\Rightarrow \] \[(5y+2)(5y+3)=0\]\[\Rightarrow \]\[y=\frac{-\,3}{5}\] or \[\frac{-\,2}{5}\] |
| Hence, x < y |
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