Question

if x+1/x=5/so find the value of x*2+1/x2 and x*4+ 1/x4​

Answer 1

Answer :-  
As provided in Question      

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

♦ So as we know Identity

$(a + b)^2 = a^2 + b^2 + 2ab \\\\ \implies a^2 + b^2 = (a+b)^2 - 2ab$


♦ Then the value of

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

>> Let us consider

$x^2$ as "a" and $\dfrac{1}{x^2}$ as "b" .


♦ Then from the value of $a^2 + b^2$ mentioned above .

$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2x\dfrac{1}{x}$
♦ By using the value of $x + \dfrac{1}{x} = 5$

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

$\implies x^2 + \dfrac{1}{x^2} = 25 - 2$

$\implies x^2 + \dfrac{1}{x^2} = 23$


♦ Now the value of

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

>> Let us consider

$x^4$ as "a" and $\dfrac{1}{x^4}$ as "b" .


♦ Then from the value of $a^2 + b^2$ mentioned above .

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

♦ By using the value of $x^2 + \dfrac{1}{x^2} = 23$ from above's answer.

$\implies x^4 + \dfrac{1}{x^4} = (23)^2 - 2$

$\implies x^4 + \dfrac{1}{x^4} = 529 - 2$

$\implies x^4 + \dfrac{1}{x^4} = 527$

Answer 2

Given

$x + \frac{1}{x} = 5$

As We know

(a+b)²=a²+b²+2ab

Using This Identity

Square Both Sides

$( x + \frac{1}{x} ) {}^{2} = 5 {}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{x} \times \frac{1}{\cancel{x}} = 25 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 25 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 23$

Now Squaring Both Sides of x²+1/x²=23

$({x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = {23}^{2} \\ \\ {x}^{2 \times 2} + \frac{1}{ {x}^{2 \times 2} } + 2 \times \cancel{x}^{2} \times \frac{1}{\cancel {x}^{2} } = 529 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 529 - 2 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 527$

$\boxed{\mathbf{\huge{{x}^{2}+\frac{1}{{x}^{2}}=23}}}$

$\boxed{\mathbf{\huge{{x}^{4}+\frac{1}{{x}^{4}}=527}}}$