Question

$\huge \: if \: x - \frac{1}{x} = \sqrt{5}$
$find \: the \: value \: of - -$
$\huge \ {x}^{4} + \frac{1}{ {x}^{4} }$
$\\ \\ \\ please \: solve$
​

Answer 1

Answer:

The required value of x⁴ + 1/x⁴ is 47.

Step-by-step explanation:

Given :

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

To find :

the value of  $\sf x^4+\dfrac{1}{x^4}$

Solution :

 $\sf x-\dfrac{1}{x}=\sqrt{5}$

Squaring both sides,

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

Applying (a - b)² = a² + b² - 2ab

$\implies \sf x^2+\dfrac{1}{x^2}-2(x)\bigg(\dfrac{1}{x}\bigg)=5 \\\\ \implies \sf x^2+\dfrac{1}{x^2}-2=5 \\\\ \implies \sf x^2+\dfrac{1}{x^2}=5+2 \\\\ \implies \sf x^2+\dfrac{1}{x^2}=7$

Squaring both sides again,

$\implies \sf \bigg(x^2+\dfrac{1}{x^2}\bigg)^2=7^2 \\\\ \implies \sf \bigg(x^2+\dfrac{1}{x^2}\bigg)^2=49$

Applying (a + b)² = a² + b² + 2ab

$\implies \sf (x^2)^2+\bigg(\dfrac{1}{x^2}\bigg)^2+2(x^2)\bigg(\dfrac{1}{x^2}\bigg)=49 \\\\ \implies \sf x^4+\dfrac{1}{x^4}+2=49 \\\\ \implies \sf x^4+\dfrac{1}{x^4}=49-2 \\\\ \implies \sf x^4+\dfrac{1}{x^4}=47$

The value of x⁴ + 1/x⁴ is 47