Question

If x⁴+1/x⁴=47. Then find the value of (x+1/x)?

Answer 1

Answer:

3 (or) -3

Step-by-step explanation:

Given :

To find the value of :

$x+\frac{1}{x}$ if, $x^4+\frac{1}{x^4} = 47$

Solution :

We know that,

$(x^2 +\frac{1}{x^2} )^2= (x^2)^2 + 2(x^2)(\frac{1}{x^2}) + (\frac{1}{x^2})^2$

⇒ $(x^2 +\frac{1}{x^2} )^2= (x^2)^2 + 2(1) + (\frac{1}{x^2})^2$

⇒ $(x^2 +\frac{1}{x^2} )^2= x^4 + 2 + \frac{1}{x^4}$

⇒ $(x^2 +\frac{1}{x^2} )^2= x^4 + \frac{1}{x^4} + 2$

⇒ $(x^2 +\frac{1}{x^2} )^2= 47 + 2$

⇒ $(x^2 +\frac{1}{x^2} )^2= 49$

⇒ $x^2 +\frac{1}{x^2} = 7$  (by taking square root both the sides )

By adding 2 both the sides, we get :

⇒ $x^2 +\frac{1}{x^2} + 2 = 7 + 2$

⇒$(x)^2 + (\frac{1}{x})^2 + 2(x)(\frac{1}{x}) = 9$

⇒ $(x +\frac{1}{x})^2 = (3)^2$

⇒ $x +\frac{1}{x}= 3(or)-3$ (by taking square root both the sides )