Question

The value of ( 1/x) + x if x = 2 -√5

Answer 1

★Given:-

$\sf x = 2 - \sqrt{5}$

★To find:-

$\sf x + \dfrac{1}{x}$

★Solution

$\sf x = 2 - \sqrt{5}$

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

Rationalising it, we will get,

$\sf \dfrac{1}{x} = \dfrac{1}{2-\sqrt{5}} \times \dfrac{2 + \sqrt{5}}{2+ \sqrt{5}}=\dfrac{2+\sqrt{5}}{(2)^2 - (\sqrt{5} )^2 }$

$\sf \implies \dfrac{1}{x} = \dfrac{2+\sqrt{5}}{4-5} = \dfrac{2+\sqrt{5}}{-1} = -2-\sqrt{5}$

______________

Now we will find,

$\sf x + \dfrac{1}{x}$

We will use the identity (a+b)²=a²+b²+2ab to find it

$\sf \implies \Bigg ( x + \dfrac{1}{x} \Bigg )^2 = (x)^2 + \Bigg ( \dfrac{1}{x} \Bigg )^2+ 2 \times x \times \dfrac{1}{x}$

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

$\sf \implies \Bigg ( x + \dfrac{1}{x} \Bigg )^2 = ( (2)^2 + (\sqrt{5})^2 - 2 \times 2 \times \sqrt{5} )+ ( (-2)^2 + (\sqrt{5} )^2 + 2 ) + 2$

$\sf \implies \Bigg ( x + \dfrac{1}{x} \Bigg )^2 = ( 4 + 5 - 4 \sqrt{5} )+ ( 4 + 5 + 4 \sqrt{5} ) + 2$

$\sf \implies \Bigg ( x + \dfrac{1}{x} \Bigg )^2 = 4 + 5 + 4 + 5+2$

$\sf \implies \Bigg ( x + \dfrac{1}{x} \Bigg )^2 =20$

$\sf \implies x + \dfrac{1}{x}=\sqrt{20}$

Answer = √20.

Answer 2

Answer:

Here is your answer

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