Question

if x=√5+2 then find x-1/x​

Answer 1

Answer:

$\ \frac{ \sqrt{5} + 2 - 1 }{ \sqrt{5 + 2} }$

$\frac{ \sqrt{5} + 1 }{ \sqrt{5} + 2 }$

$5 + 2 \sqrt{5} - \sqrt{5} - 2 \frac{}{}$

$3 \sqrt{5}$

Answer 2

Question:

If $x= \sqrt{5}+2$ then find $x - \dfrac{1}{x}$

To find:

★ To find the value of $x$ .

Answer:

$\underline{ \: \bold{ x - \frac{1}{x} = 4}\:} \: is \: the \: answer.$

Given:

$x= \sqrt{5}+2$

Step-by-step explanation:

$\bigstar \: x= \sqrt{5}+2$

$\hookrightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{5}+2}$

Now rationalising the denominator,

$\hookrightarrow \dfrac{ \sqrt{5} - 2 }{( \sqrt{5} + 2)( \sqrt{5} - 2) }$

It's of the form ${a}^{2} - {b}^{2}$

$\hookrightarrow \dfrac{ \sqrt{5} - 2 }{5 - 4}$

$\hookrightarrow \boxed{ \dfrac{1}{x} = \sqrt{5} - 2 }$

Now finding the value of $x-\dfrac{1}{x}$

$\implies \: x - \dfrac{1}{x} = \sqrt{5} + 2 - ( \sqrt{5} - 2)$

$\implies \: x - \dfrac{1}{x} = \sqrt{\not{5}} + 2 - \sqrt{\not{5}} + 2$

$\implies \: \boxed{ x - \dfrac{1}{x} = 4}$

Formula used:

$\bigstar a^{2}-b^{2}=(a+b)(a-b)$