| Directions: In each of the following questions two equations I and II are given. You have to solve both the equations and find out values of x and y and give answer. |
| I. \[6{{x}^{2}}+25x+24=0\] |
| II. \[12{{y}^{2}}+13y+3=0\] |
A) If \[x>y\]
B) If \[x\le y\]
C) If \[x<y\]
D) If relationship between x and y cannot be determined
E) If \[x\ge y\]
Correct Answer: C
Solution :
| I. \[6{{x}^{2}}+25x+24=0\] |
| \[\Rightarrow \] \[6{{x}^{2}}+16x+9x+24=0\] |
| [by splitting the middle term] |
| \[\Rightarrow \] \[2x\,(3x+8)+3\,(3x+8)=0\] |
| \[\Rightarrow \] \[(2x+3)(3x+8)=0\] |
| \[\Rightarrow \] \[x=\frac{-\,3}{2}\]and \[x=\frac{-\,8}{3}\] |
| II. \[12{{y}^{2}}+13y+3=0\] |
| \[\Rightarrow \]\[12{{y}^{2}}+9y+4y+3=0\] |
| [by splitting the middle term] |
| \[\Rightarrow \] \[3y\,(4y+3)+1\,(4y+3)=0\] |
| \[\Rightarrow \] \[(4y+3)(3y+1)=0\] |
| \[\Rightarrow \] \[y=\frac{-\,3}{4}\] and \[y=\frac{-1}{3}\] |
| \[\therefore \] \[x<y\] |
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