Question

if x plus one upon x is equal to root 5 find the value of x to the power 4 + 1 upon x to the power 4​

Answer 1

Answer:

7

Step-by-step explanation:

Explanation is written in the image linked above.

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{{x}^{4} + \dfrac{1}{ {x}^{4}} = 7}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

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

To find :- $\sf{x^4 + \dfrac{1}{x^4}}$

Solution :-

$\tt{x + \dfrac{1}{x} = \sqrt{5}}$

By squaring on both the sides :-

$\tt{{(x + \dfrac{1}{x})}^2 = (\sqrt{5})^2}$

We know that (a + b)² = a² + b² + 2ab

Here a = x, b = 1/x

By substituting the values in the identity we have :-

$\tt{{x}^{2} + \dfrac{1}{ {x}^{2} } + 2(x)( \dfrac{1}{x}) =5}$

$\tt{{x}^{2} + \dfrac{1}{ {x}^{2} } + 2 =5}$

$\tt{{x}^{2} + \dfrac{1}{ {x}^{2}} = 5 - 2}$

$\tt{{x}^{2} + \dfrac{1}{ {x}^{2}} = 3}$

By squaring on both the sides :-

$\tt{{({x}^{2} + \dfrac{1}{ {x}^{2}})}^2 = 3^2}$

We know that (a + b)² = a² + b² + 2ab

Here a = x², b = 1/x²

By substituting the values in the identity we have :-

$\tt{{({x}^{2})}^{2} + {(\dfrac{1}{ {x^2}})}^{2} + 2( {x}^{2})( \dfrac{1}{ {x}^{2} }) =9}$

$\tt{{x}^{2 \times 2} + \dfrac{1}{ {x}^{2 \times 2}} + 2=9}$

$\tt{{x}^{4} + \dfrac{1}{ {x}^{4}} = 9 - 2}$

$\tt{{x}^{4} + \dfrac{1}{ {x}^{4}} = 7}$

$\boxed{\bf{{x}^{4} + \dfrac{1}{ {x}^{4}} = 7}}$

$\mathfrak{\large{\underline{\underline{Identity\:Used:-}}}}$

(a + b)² = a² + b² + 2ab

$\mathfrak{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (a + b)² = a² + b² + 2ab

[2] (a - b)² = a² + b² - 2ab

[3] (a + b)(a - b) = a² - b²

[4] (x + a)(x + b) = x² + (a + b)x + ab