Question

# Find the value of Expression $x + \dfrac{1}{x}$ if $x = 2\sqrt 6 + 5$

Hint: Put the value of $x$ directly in the given expression and find out its value.cylindrical portion of the pillar is conical
Given, $x = 2\sqrt 6 + 5$ and we have to find the value of $x + \dfrac{1}{x}$.
Let $x + \dfrac{1}{x} = y,$then we have:
$\Rightarrow y = \dfrac{{{x^2} + 1}}{x}$
Now, putting the value of$x$, we’ll get:
$\Rightarrow y = \dfrac{{{{(2\sqrt 6 + 5)}^2} + 1}}{{(2\sqrt 6 + 5)}}$
We know that${(a + b)^2} = {a^2} + {b^2} + 2ab$, using this formula we’ll get:
$\Rightarrow y = \dfrac{{\left[ {{{(2\sqrt 6 )}^2} + {5^2} + 2 \times 2\sqrt 6 \times 5} \right] + 1}}{{(2\sqrt 6 + 5)}}, \\ \Rightarrow y = \dfrac{{24 + 25 + 20\sqrt 6 + 1}}{{(2\sqrt 6 + 5)}}, \\ \Rightarrow y = \dfrac{{50 + 20\sqrt 6 }}{{(2\sqrt 6 + 5)}}, \\ \Rightarrow y = \dfrac{{10\left( {2\sqrt 6 + 5} \right)}}{{(2\sqrt 6 + 5)}}, \\ \Rightarrow y = 10 \\$
Thus, the required value of $x + \dfrac{1}{x}$ is $10$.