Question

If x= 3+2√3 find the value of x²+1/x² and x⁴+1/x⁴​

Answer 1

Given:

The value of x -

• $\bf x = 3+2\sqrt{3}$

What To Find:

We have to find the value of -

• $\bf x^2 + \dfrac{1}{x^2}$

• $\bf x^4 + \dfrac{1}{x^4}$

Solution:

• Finding the value of -

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

We know that 1/x is the conjugate of x. That is if -

$\sf \to x = 3+2\sqrt{3}$

Then,

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

• Finding the value of -

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

Substitute the values,

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

Remove the brackets,

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

Rearrange the terms,

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

Solve the terms,

$\sf \to 6$

• Finding the value of -

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

Can be written as,

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

Write the value of sum in RHS,

$\sf \to \bigg(x + \dfrac{1}{x} \bigg)^2 = 6^2$

Use the identity (a + b)² = a² + b² + 2ab,

$\sf \to x^2 + \dfrac{1}{x^2} + 2 \bigg(x \times \dfrac{1}{x} \bigg) = 6^2$

Solve the brackets,

$\sf \to x^2 + \dfrac{1}{x^2} + 2 = 6^2$

Take 2 to RHS,

$\sf \to x^2 + \dfrac{1}{x^2} = 6^2 - 2$

Find the square of 6,

$\sf \to x^2 + \dfrac{1}{x^2} = 36 - 2$

Subtract 2 from 36,

$\sf \to x^2 + \dfrac{1}{x^2} = 34$

• Finding the value of -

$\sf \to x^4 + \dfrac{1}{x^4}$

Can be written as,

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

Also written as,

$\sf \to \bigg( x^2 + \dfrac{1}{x^2} \bigg)^2 = 34^2$

Using the identity (a + b)² = a² + b² + 2ab,

$\sf \to x^4 + \dfrac{1}{x^4} + 2\bigg(x \times \dfrac{1}{x} \bigg) = 34^2$

Solve the brackets,

$\sf \to x^4 + \dfrac{1}{x^4} + 2= 34^2$

Take 2 to RHS,

$\sf \to x^4 + \dfrac{1}{x^4}= 34^2 - 2$

Find the square of 34,

$\sf \to x^4 + \dfrac{1}{x^4}= 1156 - 2$

Subtract 2 from 1156,

$\sf \to x^4 + \dfrac{1}{x^4}= 1154$

Final Answer:

∴ Thus, the value of -

• $\bf x^2 + \dfrac{1}{x^2} \: is \: 34$

• $\bf x^4 + \dfrac{1}{x^4} \: is \: 1154$