| Directions: In the given questions, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer. |
| I. \[12{{x}^{2}}-x-1=0\] |
| II. \[20{{y}^{2}}-41y+20=0\] |
A) \[x>y\]
B) \[x\ge y\]
C) \[x<y\]
D) Relationship between a; and y cannot be determined
E) \[x\le y\]
Correct Answer: C
Solution :
| I. \[12{{x}^{2}}-x-1=0\] |
| \[D=\sqrt{{{b}^{2}}-4ac}=\sqrt{1-4\times 12\times (-1)}\] |
| \[=\sqrt{1+48}=\sqrt{49}=7\] |
| \[{{x}_{1}}=\frac{-\,b+D}{2a}=\frac{1+7}{24}=\frac{8}{24}=\frac{1}{3}\] |
| \[{{x}_{2}}=\frac{-\,b-D}{2a}\] |
| \[\Rightarrow \] \[{{x}^{2}}=\frac{1-7}{24}=\frac{-\,6}{24}=\frac{-1}{4}\] |
| \[\Rightarrow \] \[x=\frac{1}{3},\] \[-\,\,\frac{1}{4}\] |
| II. \[20{{y}^{2}}-41y+20\] |
| \[{{y}_{1}}=\frac{41+\sqrt{1681-1600}}{40}\] |
| \[\Rightarrow \] \[{{y}_{1}}=\frac{41+9}{40}=\frac{50}{40}=\frac{5}{4}\] |
| and \[{{y}_{2}}=\frac{41-\sqrt{1681-1600}}{40}\] |
| \[\Rightarrow \] \[{{y}_{2}}=\frac{32}{40}=\frac{4}{5}\]\[\Rightarrow \]\[y=\frac{5}{4},\]\[\frac{4}{5}\] |
| \[\therefore \] \[x<y\] |
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