| Directions: In the given questions, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer. |
| I. \[6{{x}^{2}}+25x+24=0\] |
| II. \[12{{y}^{2}}+13y+3=0\] |
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. \[6{{x}^{2}}+25x+24=0\] |
| \[D=\sqrt{{{b}^{2}}-4ac}\] |
| \[\Rightarrow \] \[D=\sqrt{625-4\times 24\times 6}=\sqrt{49}=7\] |
| \[{{x}_{1}}=\frac{-\,b+7}{12}=\frac{-\,25+7}{12}=\frac{-18}{12}=-\,\,\frac{3}{2}\] |
| \[{{x}_{2}}=\frac{-\,b-7}{12}=\frac{-\,25-7}{12}=\frac{-\,32}{12}=-\,\,\frac{8}{3}\] |
| \[\Rightarrow \] \[x=\frac{-\,3}{2},\,\]\[\frac{-\,8}{3}\] |
| II. \[12{{y}^{2}}+13y+3=0\] |
| \[{{y}_{1}}=\frac{-13+\sqrt{169-144}}{24}\] |
| \[=\frac{-13+5}{24}=\frac{-\,8}{24}=\frac{-1}{3}\] |
| \[{{y}_{2}}=\frac{-13-\sqrt{169-144}}{24}\] |
| \[=\frac{-13-5}{24}=\frac{-18}{24}=\frac{-\,3}{4}\] |
| \[\Rightarrow \] \[y=\frac{-1}{3},\]\[\frac{-\,3}{4}\] |
| \[\therefore \] \[x<y\] |
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