Question

Date
Pago
is
If x= √5 +2, then an
equal
x -1/x​

Answer 1

Given:

The value of x -

• $\bf \to x = \sqrt{5} + 2$

What To Find:

We have to find -

• The value of $\bf \to x - \dfrac{1}{x}$

Solution:

• Finding the value of -

$\bf \to \dfrac{1}{x}$

We know that -

$\bf \to x = \sqrt{5} + 2$

Substitute the value,

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

Rationalise the denominator with its conjugate,

$\bf \to \dfrac{1}{x} = \dfrac{1}{\sqrt{5} + 2} \times \dfrac{\sqrt{5} - 2}{\sqrt{5} - 2}$

Take them as common,

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

Solve the numerator,

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

Solve the denominator by using the identity (a +  b) (a - b) = a² - b²,

$\bf \to \dfrac{1}{x} = \dfrac{\sqrt{5} - 2}{(\sqrt{5})^2 - (2)^2}$

Find the squares,

$\bf \to \dfrac{1}{x} = \dfrac{\sqrt{5} - 2}{5 - 4}$

Subtract the values,

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

Can be written as,

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

• Finding the value of -

$\bf \to x - \dfrac{1}{x}$

Substitute the values,

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

Remove the brackets,

$\bf \to x - \dfrac{1}{x} =\sqrt{5} + 2 - \sqrt{5} + 2$

Rearrange the terms,

$\bf \to x - \dfrac{1}{x} =\underline{\sqrt{5} - \sqrt{5}} + \underline{2 + 2}$

Solve the first term,

$\bf \to x - \dfrac{1}{x} =0 + \underline{2 + 2}$

Solve the second term,

$\bf \to x - \dfrac{1}{x} =0 + 4$

Can be written as,

$\bf \to x - \dfrac{1}{x} = 4$

Final Answer:

∴ Thus, the value of $\bf x - \dfrac{1}{x}$ is $\bf 4$.