Question

find x+1/x if x=4-√5​

Answer 1

Given :

$\rm x = 4-\sqrt{5}\qquad ..[1]$

Hence,

$\rm\dashrightarrow\dfrac{1}{x} = \dfrac{1}{4-\sqrt{5}}$

Rationalising the denominator.

We rationalise a denominator by multiplying the conjugate of the denominator with numerator and denominator.

Here, conjugate of the denominator is 4 + √5

$\rm\dashrightarrow\dfrac{1}{4-\sqrt{5}} \times \dfrac{4+\sqrt{5}}{4+\sqrt{5}}$

$\rm\dashrightarrow \dfrac{4+\sqrt{5}}{(4+\sqrt{5})(4-\sqrt{5})}$

We know that,

a² - b² = (a + b)(a - b)

$\rm\dashrightarrow \dfrac{4+\sqrt{5}}{(4)^2-(\sqrt{5})^2}$

$\rm\dashrightarrow \dfrac{4+\sqrt{5}}{16-5}$

$\rm\dashrightarrow \dfrac{4+\sqrt{5}}{11} \qquad ...[2]$

Now we will find x + 1/x

Substituting the values of x and 1/x from [1] and [2]

$\rm\dashrightarrow x+\dfrac{1}{x} = 4-\sqrt{5} + \dfrac{4+\sqrt{5}}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{11(4-\sqrt{5})}{11} + \dfrac{4+\sqrt{5}}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{44-11\sqrt{5}}{11} + \dfrac{4+\sqrt{5}}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{44-11\sqrt{5}+(4+\sqrt{5})}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{44-11\sqrt{5}+4+\sqrt{5}}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{44+4-11\sqrt{5}+\sqrt{5}}{11}$

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{48-10\sqrt{5}}{11}$

You can simplify it further if you want.

$\rm\dashrightarrow x+\dfrac{1}{x} = \dfrac{2(24-5\sqrt{5})}{11}$

$\bf\dashrightarrow x+\dfrac{1}{x} = \dfrac{2}{11}(24-5\sqrt{5})$