| Directions: In the following questions, two equations numbered I and II have been given. You have to solve both the equations and mark the correct answer. |
| I. \[3{{x}^{2}}+29x+56=0\] |
| II. \[2{{y}^{2}}+15y+25=0\] |
A) If \[x<y\]
B) If \[x>y\]
C) If \[x\ge y\]
D) If \[x\le y\]
E) If relationship between \[x\] and \[y\] cannot be established
Correct Answer: A
Solution :
| I. \[3{{x}^{2}}+29x+56=0\] |
| \[\Rightarrow \]\[3{{x}^{2}}+21x+8x+56=0\] |
| \[\Rightarrow \]\[3x\,\,(x+7)+8\,\,(x+7)=0\] |
| \[\Rightarrow \] \[(3x+8)(x+7)=0\] \[\Rightarrow \] \[x=-\,\,7,\]\[x=-\frac{8}{3}\] |
| II. \[2{{y}^{2}}+15y+25=0\] |
| \[\Rightarrow \]\[2{{y}^{2}}+10y+5y+25=0\] |
| \[\Rightarrow \]\[2y\,\,(y+5)+5\,\,(y+5)=0\] |
| \[\Rightarrow \] \[(y+5)(2y+5)=0\] |
| \[\Rightarrow \] \[y=-\,\,5,\]\[y=-\,\,5/2\] |
| \[\therefore \]\[x<y\] |
You need to login to perform this action.
You will be redirected in
3 sec
