Question

the solution set of (x2+1/x2)-5(x+1/x) +6=0 is​

Answer 1

We have to find the solution set of the equation,

$\longrightarrow\sf{\left(x^2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)+6=0\quad\quad\dots(1)}$

Let,

$\longrightarrow\sf{x+\dfrac{1}{x}=k}$

On squaring both sides,

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

$\longrightarrow\sf{x^2+2\cdot x\cdot\dfrac{1}{x}+\dfrac{1}{x^2}=k^2}$

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

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

Then (1) becomes,

$\longrightarrow\sf{\left(k^2-2\right)-5k+6=0}$

$\longrightarrow\sf{k^2-5k+4=0}$

$\longrightarrow\sf{k^2-k-4k+4=0}$

$\longrightarrow\sf{k(k-1)-4(k-1)=0}$

$\longrightarrow\sf{(k-1)(k-4)=0}$

$\Longrightarrow\sf{k=1\quad OR\quad k=4}$

$\longrightarrow\sf{x+\dfrac{1}{x}=1\quad OR\quad x+\dfrac{1}{x}=4}$

Case 1:- Let $\sf{x+\dfrac{1}{x}=1.}$

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

$\longrightarrow\sf{x^2+1=x}$

$\longrightarrow\sf{x^2-x+1=0}$

$\Longrightarrow\sf{x=\dfrac{1\pm\sqrt{(-1)^2-4\times1\times1}}{2\times1}}$

$\longrightarrow\sf{x=\dfrac{1\pm\sqrt{1-4}}{2}}$

$\longrightarrow\sf{x=\dfrac{1\pm\sqrt{-3}}{2}}$

$\longrightarrow\sf{x=\dfrac{1\pm i\sqrt{3}}{2}}$

Case 2:- Let $\sf{x+\dfrac{1}{x}=4.}$

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

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

$\longrightarrow\sf{x^2-4x+1=0}$

$\Longrightarrow\sf{x=\dfrac{4\pm\sqrt{(-4)^2-4\times1\times1}}{2\times1}}$

$\longrightarrow\sf{x=\dfrac{4\pm\sqrt{16-4}}{2}}$

$\longrightarrow\sf{x=\dfrac{4\pm\sqrt{12}}{2}}$

$\longrightarrow\sf{x=\dfrac{4\pm2\sqrt{3}}{2}}$

$\longrightarrow\sf{x=2\pm\sqrt{3}}$

Hence the solution set is,

$\longrightarrow\textsf{ \underline{\underline{\mathsf{x\in\left\{\dfrac{1\pm i\sqrt3}{2},\ 2\pm\sqrt3\right\}}}} }$

For $\sf{x\in\mathbb{R},}$

$\longrightarrow\underline{\underline{\sf{x\in\left\{2\pm\sqrt3\right\}}}}$

Answer 2

Answer:

(x^2+1/x^2)-5(x+1/x)+6=0