Question

If x^4+1/x^4=623 then find the value off+1/x

Answer 1

Given,

$\longrightarrow x^4+\dfrac{1}{x^4}=623$

Add 2 to both sides.

$\longrightarrow x^4+\dfrac{1}{x^4}+2=623+2$

$\longrightarrow (x^2)^2+\left(\dfrac{1}{x^2}\right)^2+2\times x^2\times\dfrac{1}{x^2}=625$

$\longrightarrow \left(x^2+\dfrac{1}{x^2}\right)^2=25^2$

Let,

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

Adding 2 to both,

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

$\longrightarrow x^2+\left(\dfrac{1}{x}\right)^2+2\times x\times\dfrac{1}{x}=27$

$\longrightarrow \left(x+\dfrac{1}{x}\right)^2=(3\sqrt3)^2$

$\longrightarrow\underline{\underline{x+\dfrac{1}{x}=\pm3\sqrt3}}$

Let,

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

Adding 2 to both,

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

$\longrightarrow x^2+\left(\dfrac{1}{x}\right)^2+2\times x\times\dfrac{1}{x}=-23$

$\longrightarrow \left(x+\dfrac{1}{x}\right)^2=-23$

$\longrightarrow\underline{\underline{x+\dfrac{1}{x}=\pm i\sqrt{23}}}$