| If \[(a+b):\sqrt{ab}=4:1\]where \[a>b>0,\]then \[a:b\]is equal to |
A) \[(2+\sqrt{3}):(2-\sqrt{3})\]
B) \[(2-\sqrt{3}):(2+\sqrt{3})\]
C) \[(3+\sqrt{2}):(3-\sqrt{2})\]
D) \[(3-\sqrt{2}):(3+\sqrt{2})\]
Correct Answer: A
Solution :
| \[\frac{a+b}{\sqrt{ab}}=\frac{4}{1}\]\[\Rightarrow \]\[\frac{a+b}{2\sqrt{ab}}=\frac{2}{1}\] |
| \[\Rightarrow \]\[\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}\] |
| [by componendo and dividendo] |
| \[\Rightarrow \] \[\frac{{{(\sqrt{a})}^{2}}+{{(\sqrt{b})}^{2}}+2\sqrt{a}\sqrt{b}}{{{(\sqrt{a})}^{2}}+{{(\sqrt{b})}^{2}}-2\sqrt{a}\sqrt{b}}=\frac{3}{1}\] |
| \[\Rightarrow \]\[\frac{{{(\sqrt{a}+\sqrt{b})}^{2}}}{{{(\sqrt{a}-\sqrt{b})}^{2}}}=\frac{3}{1}\]\[\Rightarrow \]\[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}\] |
| \[\Rightarrow \] \[\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\] |
| [by componendo and dividendo] |
| \[\Rightarrow \] \[\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\] |
| \[\Rightarrow \] \[\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\] |
| On squaring both sides, we get |
| \[{{\left( \frac{\sqrt{a}}{\sqrt{b}} \right)}^{2}}=\frac{{{(\sqrt{3}+1)}^{2}}}{{{(\sqrt{3}-1)}^{2}}}\] |
| \[\Rightarrow \] \[\frac{a}{b}=\frac{{{(\sqrt{3})}^{2}}+{{(1)}^{2}}+2\sqrt{3}}{{{(\sqrt{3})}^{2}}+{{(1)}^{2}}-2\sqrt{3}}\]\[\Rightarrow \]\[\frac{a}{b}=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}\] |
| \[\Rightarrow \] \[\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\] |
| \[\therefore \] \[a:b=(2+\sqrt{3}):(2-\sqrt{3})\] |
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