| If \[n=7+4\sqrt{3},\]then the value of \[\left( \sqrt{n}+\frac{1}{\sqrt{n}} \right)\] is [SSC (CGL) 2012] |
A) \[2\sqrt{3}\]
B) \[4\]
C) \[-\,\,4\]
D) \[-\,\,2\sqrt{3}\]
Correct Answer: B
Solution :
| \[n=7+4\sqrt{3}=7+2\times 2\times \sqrt{3}\] |
| \[=4+3+2\times 2\times \sqrt{3}={{(2)}^{2}}+{{(\sqrt{3})}^{2}}+2\times 2\times \sqrt{3}\] |
| \[={{(2+\sqrt{3})}^{2}}\] \[[\because {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] |
| Taking square root on both sides, we get. |
| \[\sqrt{n}=2+\sqrt{3}\] ... (i) |
| \[\therefore \] \[\frac{1}{\sqrt{n}}=\frac{1}{2+\sqrt{3}}\] [reciprocal] |
| \[=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}\] [rationatising] |
| \[=2-\sqrt{3}\] ... (ii) |
| On adding Eqs. (i) and (ii), we get |
| \[\sqrt{n}+\frac{1}{\sqrt{n}}=2+\sqrt{3}+2-\sqrt{3}=4\] |
| Alternate Method |
| Given, \[n=7+4\sqrt{3}\] |
| \[\therefore \]\[\frac{1}{n}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}\] |
| [rationalising] |
| \[=\frac{7-4\sqrt{3}}{{{(7)}^{2}}-{{(4\sqrt{3})}^{2}}}\]\[[\because (a+b)(a-b)={{a}^{2}}-{{b}^{2}}]\] |
| \[=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}\] |
| We know that, |
| \[{{\left( \sqrt{n}+\frac{1}{\sqrt{n}} \right)}^{2}}=\left( n+\frac{1}{n}+2\times \sqrt{n}\times \frac{1}{\sqrt{n}} \right)\] |
| \[\Rightarrow \] \[\left( \sqrt{n}+\frac{1}{\sqrt{n}} \right)=\sqrt{n+\frac{1}{n}+2}\] |
| \[=\sqrt{7+4\sqrt{3}+7-4\sqrt{3}+2}\] |
| \[=\sqrt{14+2}=\sqrt{16}=4\] |
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