A) \[3n{{x}^{2}}-2mx+3n=0\]
B) \[2n{{x}^{2}}-2mx+3n=0\]
C) \[3n{{x}^{2}}-2mx-3n=0\]
D) \[3n{{x}^{2}}+3mx+3n=0\]
Correct Answer: A
Solution :
| \[\frac{x}{1}=\frac{\sqrt{m+3n}+\sqrt{m-3n}}{\sqrt{m+3n}-\sqrt{m-3n}}\] |
| By componendo and dividendo rule, |
| \[\frac{x+1}{x-1}=\frac{\sqrt{m+3n}+\sqrt{m-3n}+\sqrt{m+3n}-\sqrt{m-3n}}{\sqrt{m+3n}+\sqrt{m-3n}-\sqrt{m+3n}+\sqrt{m-3n}}\] |
| \[\Rightarrow \]\[\frac{x+1}{x-1}=\frac{\sqrt{m+3n}}{\sqrt{m-3n}}\] |
| Squaring both sides, |
| \[\frac{{{(x+1)}^{2}}}{{{(x-1)}^{2}}}=\frac{m+3n}{m-3n}\] |
| Applying componendo and dividendo rule, |
| \[\frac{{{(x+1)}^{2}}+{{(x-1)}^{2}}}{{{(x+1)}^{2}}-{{(x-1)}^{2}}}=\frac{m+3n+m-3n}{m+3n-m+3n}\] |
| \[\Rightarrow \]\[\frac{2\,\,({{x}^{2}}+1)}{2\,\,(2x)}=\frac{2m}{2\,\,(3n)}\] |
| \[\Rightarrow \]\[\frac{{{x}^{2}}+1}{2x}=\frac{m}{3n}\] |
| \[\Rightarrow \]\[3n{{x}^{2}}-2mx+3n=0\] |
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