| Directions: In each of the following questions there are two equations I and II. You have to solve both equations and give answer. [SBI (PO) 2015] |
| I. \[14{{x}^{2}}-13x+3=0\] |
| II. \[99{{y}^{2}}-62y+8=0\] |
A) If \[x>y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x\le y\]
E) If \[x=y\]or relation cannot be established
Correct Answer: A
Solution :
| I. \[14{{x}^{2}}-13x+3=0\] |
| \[\Rightarrow \] \[14{{x}^{2}}-6x-7x+3=0\] |
| \[\Rightarrow \] \[2x\,(7x-3)-1\,(7x-3)=0\]\[\Rightarrow \,\]\[x=\frac{1}{2},\]\[\frac{3}{7}\] |
| II. \[99{{y}^{2}}-62y+8=0\] |
| \[\Rightarrow \] \[99{{y}^{2}}-44y-18y+8=0\] |
| \[\Rightarrow \] \[11y\,(9y-4)-2\,(9y-4)=0\]\[\Rightarrow \,\]\[y=\frac{2}{11},\]\[\frac{4}{9}\] |
| \[\therefore \] \[x>y\] |
You need to login to perform this action.
You will be redirected in
3 sec
