Question

2x^4-7x^3+9x^2-7x+2=0
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Answer 1

$2x^4-7x^3+9x^2-7x+2=0\\\\(2x^4+2)-(7x^3+7x)+9x^2=0\; |:x^2\ne 0\\\\2\, (x^2+\dfrac{1}{x^2})-7(x+\dfrac{1}{x})+9=0\\\\t=x+\dfrac{1}{x}\ \ ,\ \ \ t^2=x^2+2+\dfrac{1}{x^2}\ \ ,\ \ x^2+\dfrac{1}{x^2}=t^2-2\\\\2\, (t^2-2)-7t+9=0\ \ \ \to \ \ \ 2t^2-7t+5=0 \ \,\ \ ,\ \ D=9\\\\t_1=1\ \ ,\ \ t_2=\dfrac{5}{2}\\\\a)\ \ x+\dfrac{1}{x}=1\ \ ,\ \ \dfrac{x^2-x+1}{x}=0\ \ ,\ \ x^2-x+1=0\ \ ,\ \ D=-3<0\ \ \to \ \ x\in \varnothing \\\\b)\ \ x+\dfrac{1}{x}=\dfrac{5}{2}\ \ ,\ \ \dfrac{2x^2-5x+2}{2x}=0\ \ ,\ \ 2x^2-5x=2=0\ ,$

$D=9\ \ ,\ \ x_1=\dfrac{1}{2}\ \ ,\ \ x_2=2\\\\Otvet:\ \ x_1=\dfrac{1}{2}\ \ ,\ \ x_2=2\; .$