Question

2(x^2+1/x^2 )-9(x+1/x)+14=0 solve

Answer 1

$2(x^2+\frac{1}{x^2} )-9(x+\frac{1}{x})+14=0$........(1)

$\text{Take, }t=x+\frac{1}{x}$

$\implies\,t^2=x^2+\frac{1}{x^2}+2$

$\implies\,x^2+\frac{1}{x^2}=t^2-2$

$\text{(1) can be written as}$

$2(t^2-2)-9t+14=0$

$2t^2-9t+10=0$

$(t-2)(2t-5)=0$

$\implies\,t=2,\frac{5}{2}$

$\textbf{when }t=2$

$\implies\,x+\frac{1}{x}=2$

$\implies\,x^2-2x+1=0$

$\implies\,(x-1)^2-=0$

$\implies\,x=1,\;1$

$\textbf{when }t=\frac{5}{2}$

$\implies\,x+\frac{1}{x}=2$

$\implies\,2x^2-5x+2=0$

$\implies\,(x-2)(2x-5)=0$

$\implies\,x=2,\;\frac{5}{2}$

$\therefore\textbf{The roots are }\bf\,1,\;1,\;2,\;\frac{5}{2}$

Answer 2

Answer:

above answer is little bit wrong at last steps

Step-by-step explanation:

x+x1=2

\implies\,2x^2-5x+2=0⟹2x2−5x+2=0

\implies\,(x-2)(2x-5)=0⟹(x−2)(2x−5)=0

\implies\,x=2,\;\frac{5}{2}⟹x=2,25