Question

If x = √3-√2 , then value of (x + 1 / x) will be​

Answer 1

Let √x - 1/√x = a

Squaring both the sides,

x + 1/x - 2 = a^2

Putting the value,

3–2√2 + 1/(3–2√2) - 2 = a^2

a^2 = 1 - 2√2 + 1/(3–2√2)

= [(3–2√2) (1–2√2) + 1] / 3–2√2

= {3 - 8√2 + 9} / 3–2√2

$= [12 - 8\sqrt{2 ] / 3–2√2$

Rationalising both the sides

= {(12 - 8√2)(3+2√2)} ÷ (9–8)

= 36 + 24√2 - 24√2 + 16(2)

= 36 - 32

=> 4

a^2 = 4

a = √4

a = 2, -2

So,

√x - 1/√x = a = 2, -2.

Answer 2

Given:

The value of x -

• $\bf x = \sqrt{3} - \sqrt{2}$

What To Find:

We have to find the value of -

• $\bf x + \dfrac{1}{x}$

Solution:

• Finding the value of -

$\sf \to \: \dfrac{1}{x}$

We know that -

$\sf \to \: x = \sqrt{3} - \sqrt{2}$

Substitute the value in 1/x,

$\sf \to \: \dfrac{1}{x} = \dfrac{1}{\sqrt{3}-\sqrt{2}}$

Let's rationalise the denominator!

Here, the rationalising factor of the denominator is -

$\sf \to \: \sqrt{3} + \sqrt{2}$

Multiply it with the expression,

$\sf \to \: \dfrac{1}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Take them as common,

$\sf \to \: \dfrac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}-\sqrt{2}) \times (\sqrt{3}+\sqrt{2})}$

Using the identity (a - b) (a + b) = a² - b²,

$\sf \to \: \dfrac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}$

Find the squares,

$\sf \to \: \dfrac{\sqrt{3}+\sqrt{2}}{3-2}$

Subtract 2 from 3,

$\sf \to \: \dfrac{\sqrt{3}+\sqrt{2}}{1}$

Also written as,

$\sf \to \: \sqrt{3}+\sqrt{2}$

∴ Thus, we got the value of 1/x.

• Finding the value of -

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

Substitute the values,

$\sf \to \: (\sqrt{3}-\sqrt{2}) + (\sqrt{3}+\sqrt{2})$

Remove the brackets,

$\sf \to \: \sqrt{3}-\sqrt{2} + \sqrt{3}+\sqrt{2}$

Rearrange the terms,

$\sf \to \: \sqrt{3}+\sqrt{3}+\sqrt{2}-\sqrt{2}$

Add √3 and √3,

$\sf \to \: 2\sqrt{3}+\sqrt{2}-\sqrt{2}$

Subtract √2 from √2,

$\sf \to \: 2\sqrt{3}+0$

Can be written as,

$\sf \to \: 2\sqrt{3}$

Final Answer:

$\rm \therefore Thus, the \: value \: of \: \bigg(x + \dfrac{1}{x} \bigg) \: is \: 2\sqrt{3}.$