Question

if x + 1/x = 7 find the value of . 1)x power 2+ 1 / x power 2. , 2)x power 4 +1/x power 4​

Answer 1

Given :-

$x +\dfrac{1}{x} = 7$

To find :-

• $x^{2} +\dfrac{1}{x^{2} }$

• $x^{4} +\dfrac{1}{x^{4} }$

Identity to use :-

(a+b)² = a² + 2ab + b²

Substituting the vales,

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

$7^{2} = x^{2} + 2(\not x)(\dfrac{1}{\not x}) + \dfrac{1}{x^{2} }$

$49 = x^{2} +2 + \dfrac{1}{x^{2} }$

$49 - 2 = x^{2} + \dfrac{1}{x^{2} }$

$47 = x^{2} + \dfrac{1}{x^{2} }$

By again using the identity,

$\left( x^2+\dfrac{1}{x^2} \right )^2=x^2+2x^2\left ( \dfrac{1}{x^2} \right )+\left ( \dfrac{1}{x^2} \right )^2$

$47^{2} = x^{4} + 2(\not x^{2} )(\dfrac{1}{\not x^{2} }) + \dfrac{1}{x^{4} }$

$2209 = x^{4} + 2 + \dfrac{1}{x^{4} }$

$2209 - 2 = x^{4} + \dfrac{1}{x^{4} }$

$2207 = x^{4} + \dfrac{1}{x^{4} }$

Answer :-

$47 = x^{2} +\dfrac{1}{x^{2} }$

$2207 = x^{4} +\dfrac{1}{x^{4}}$

Some more identities :-

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

a²-b² = (a+b)(a-b)

(x+a)(x+b) = x² + x(a+b) + ab

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca