| Directions: In the following questions two equations numbered I and II are given. You have to solve both the equation and given answer. [Bank of Baroda (PO) 2011] |
| I. \[3{{x}^{2}}-19x+28=0\] |
| II. \[5{{y}^{2}}-18y+16=0\] |
A) If \[x>y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x\le y\]
E) If \[x=y\] or the relationship cannot be established
Correct Answer: A
Solution :
| I. \[3{{x}^{2}}-19x+28=0\] |
| \[\Rightarrow \]\[3{{x}^{2}}-12x-7x+28=0\] |
| \[\Rightarrow \]\[3x\,\,(x-4)-7\,\,(x-4)=0\] |
| \[\Rightarrow \] \[(3x-7)(x-4)=0\] |
| \[\Rightarrow \] \[x=4,\]\[\frac{7}{3}\] |
| II. \[5{{y}^{2}}-18y+16=0\] |
| \[\Rightarrow \] \[5{{y}^{2}}-10y-8y+16=0\] |
| \[\Rightarrow \]\[5y\,\,(y-2)-8\,\,(y-2)=0\] |
| \[\Rightarrow \] \[(5y-8)(y-2)=0\] |
| \[\Rightarrow \] \[y=2,\]\[\frac{8}{5}\] |
| \[\therefore \] \[x>y\] |
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