if x plus one upon x is equal to root 5 find the value of x to the pow

1.	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» $\mathfrak{\large{\underline{\underline{Answer:-}}}}$ $\boxed{\bf{{x}^{<klux>4</klux>} + \dfrac{1}{ {x}^{4}} = 7}}$ $\mathfrak{\large{\underline{\underline{Explanation:-}}}}$ 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