Question

if x is equal to bracket 2 + root 3 bracket close find the value of bracket x square + 1 upon x square bracket close

Answer 1

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

By Rationalization,


$\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \\ => \frac{1}{x} = \frac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3}) }^{2} } \\ \\ \\ = > \frac{1}{x} = \frac{ 2 - \sqrt{3} }{4 - 3} \\ \\ \\ => \frac{1}{x} = 2 - \sqrt{3}$

Hence,


$x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ \\ => x + \frac{1}{x} = 4$



Square on both sides,


$(x + \frac{1}{x} ) ^{2} = {4}^{2} \\ \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ \\ => {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 14$

Answer 2

Answer:

$$\begin{lgathered}x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} }\end{lgathered}$$

By Rationalization,

$$\begin{lgathered}\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \\ => \frac{1}{x} = \frac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3}) }^{2} } \\ \\ \\ = > \frac{1}{x} = \frac{ 2 - \sqrt{3} }{4 - 3} \\ \\ \\ => \frac{1}{x} = 2 - \sqrt{3} \\\end{lgathered}$$

Hence,

$$\begin{lgathered}x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ \\ => x + \frac{1}{x} = 4\end{lgathered}$$

Square on both sides,

$$\begin{lgathered}(x + \frac{1}{x} ) ^{2} = {4}^{2} \\ \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ \\ => {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 14\end{lgathered}$$