Question

if X=5-2✓6 then find the value of x²+1/x²​

Answer 1

Answer :

Please take a look on the attachment.

Answer 2

Given:

The value of x -

• $\bf x = 5 - 2\sqrt{6}$

What To Find:

We have to find the value of -

• $\bf x^2 + \dfrac{1}{x^2}$

How To Find:

To find we have to -

• First, find the value of $\bf \dfrac{1}{x}$.
• Next, find the value of $\bf x + \dfrac{1}{x}$.
• Then, expand $\bf x^2 + \dfrac{1}{x^2}$ using identities.
• Then, solve $\bf x^2 + \dfrac{1}{x^2}$ to get the results.

Solution:

• Finding the value of -

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

We know that,

$\implies \sf x = 5 - 2\sqrt{6}$

So, it will be,

$\implies \sf \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}}$

By rationalising the denominator,

$\implies \sf \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} \times \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}}$

Take them as common,

$\implies \sf \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{5 - 2\sqrt{6} \times 5 + 2\sqrt{6}}$

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

• a = 5
• b = 2√6

$\implies \sf \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2}$

Removing the brackets,

$\implies \sf \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{5 \times 5 - 2 \times 2 \times \sqrt{6} \times \sqrt{6}}$

Multiple the terms,

$\implies \sf \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{25 - 24}$

Subtract the numbers,

$\implies \sf \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{1}$

Also written as,

$\implies \sf \dfrac{1}{x} = 5 + 2\sqrt{6}$

• Finding the value of -

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

Substitute the values,

$\implies \sf{x+\dfrac{1}{x}} = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6})$

Remove the brackets,

$\implies \sf{x+\dfrac{1}{x}} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6}$

Rearrange the terms,

$\implies \sf{x+\dfrac{1}{x}} = 5 + 5 - 2\sqrt{6} + 2\sqrt{6}$

Add 5 and 5,

$\implies \sf{x+\dfrac{1}{x}} = 10 - 2\sqrt{6} + 2\sqrt{6}$

Add -2√6 and 2√6,

$\implies \sf{x+\dfrac{1}{x}} = 10 + 0$

Also written as,

$\implies \sf{x+\dfrac{1}{x}} = 10$

• Expanding the value of -

$\implies \sf{x^2+\dfrac{1}{x^2}}$

Also written as,

$\implies \sf \bigg(x+\dfrac{1}{x}\bigg)^2$

Square both sides,

$\implies \sf \bigg(x+\dfrac{1}{x}\bigg)^2 = (10)^2$

Square the RHS,

$\implies \sf \bigg(x+\dfrac{1}{x}\bigg)^2 = 100$

Square the LHS using the identity (a + b)² = a² + b² + 2ab where,

• a = x

• b = $\sf \dfrac{1}{x}$

$\implies \sf x^2 + \dfrac{1}{x^2} + 2\bigg(x \: \times \dfrac{1}{x} \bigg) = 100$

Solve the brackets by cancelling the x's,

$\implies \sf x^2 + \dfrac{1}{x^2} + 2 = 100$

Take 2 to RHS,

$\implies \sf x^2 + \dfrac{1}{x^2} = 100 - 2$

Subtract 2 from 100,

$\implies \sf x^2 + \dfrac{1}{x^2} = 98$

Final Answer:

$\bf \therefore \: The \: value \: of \: x^2 + \dfrac{1}{x^2} \: is \: 98.$