| Directions: In these questions two equations are given. You have to solve both the equations and give answer. |
| I. \[3{{x}^{2}}-20x-32=0\] |
| II. \[2{{y}^{2}}-3y-20=0\] |
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: D
Solution :
| I. \[3{{x}^{2}}-20x-32=0\] |
| \[\Rightarrow \]\[3{{x}^{2}}-12x-8x-32=0\] |
| \[\Rightarrow \]\[3x\,\,(x-4)-8\,\,(x-4)=0\] |
| \[\Rightarrow \] \[(3x-8)(x-4)=0\] |
| \[\therefore \] \[x=\frac{8}{3},\]\[4\] |
| II. \[2{{y}^{2}}-3y-20=0\] |
| \[\Rightarrow \]\[2{{y}^{2}}-8y+5y-20=0\] |
| \[\Rightarrow \]\[2y\,\,(y-4)+5(y-4)=0\] |
| \[\Rightarrow \] \[(2y+5)(y-4)=0\] |
| \[\therefore \] \[y=4,\]\[-\frac{5}{2}\] |
| Hence, \[x\ge y\] |
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