Question

if x=2√2+√7 find x+1/x​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $x=2\sqrt{2}+\sqrt{7}$, find $x+\frac{1}{x}$.

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

$x = 2 \sqrt{2} + \sqrt{7}$

$\implies \frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} }$

$\implies \frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} } \times \frac{2 \sqrt{2} - \sqrt{7} }{2 \sqrt{2} - \sqrt{7} }$

$\implies \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{(2 \sqrt{2} )^{2} + ( \sqrt{7})^{2} }$

$\implies \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{8 - 7 }$

$\implies \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{1}$

$\implies \frac{1}{x} = 2 \sqrt{2} - \sqrt{7}$

So,

$x + \frac{1}{x} = 2 \sqrt{2} + \sqrt{7} + 2 \sqrt{2} - \sqrt{7}$

$\implies x + \frac{1}{x} = 4 \sqrt{2}$

$\star\:\:\:\bf\large\underline\blue{Answer:-}$

• $x + \frac{1}{x} = 4 \sqrt{2}$

Answer 2

$\bf\large\blue{\underline{Question}\::-}$

• $\bf{If \: x = 2 \sqrt{2} + \sqrt{7}, \: then \: find \: x + \frac{1}{x} }$

$\bf\large\blue{\underline{Solution}\::-}$

Given,

$x = 2 \sqrt{2} + \sqrt{7}$

$\therefore \: \frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} }$

$\frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} } \times \frac{2 \sqrt{2 } - \sqrt{7} }{2 \sqrt{2} - \sqrt{7} }$

$= \frac{2 \sqrt{2} - \sqrt{7} }{ {(2 \sqrt{2}) }^{2} - {( \sqrt{7} )}^{2} }$

$= \frac{2 \sqrt{2} - \sqrt{7} }{8 - 7}$

$\therefore \: \frac{1}{x} = 2 \sqrt{2} - \sqrt{7}$

$So, \: x + \frac{1}{x} = 2 \sqrt{2} + \sqrt{7} + 2 \sqrt{2} - \sqrt{7}$

$= 4 \sqrt{2}$

$\bf\large\blue{\underline{Answer}\::-}$

• The value of $x + \frac{1}{x}$ is $4 \sqrt{2}$

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$\text\blue{Hope it helps you.}$