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Reciprocal equation problems - easy

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Problem №1

Solve the reciprocal equation $x^4-3x^3+4x^2-3x+1=0$ .

Solution:
We see that it is a reciprocal equation, so we divide it by $x^2$ : $\frac{x^4}{x^2}-3\frac{x^3}{x^2}+4\frac{x^2}{x^2}-3\frac{x}{x^2}+\frac{1}{x^2}=0$
$x^2+\frac{1}{x^2}-3(x+\frac{1}{x})+4=0$ . We'll now substitute $x+\frac{1}{x}=u$ . By squaring it, we get $x^2+2x.\frac{1}{x}+\frac{1}{x^2}=u^2$ , or $x^2+\frac{1}{x^2}=u^2-2$ . We plug back into the equation to get
$u^2-2-3u+4=0$ , or $u^2-3u+2=0$ . The solutions are $u_1=1$ and $u_2=2$ . So either $x+\frac{1}{x}=1$ or $x+\frac{1}{x}=2$ . From the first solution, we get $x^2-x+1=0$ , which has no solutions. From the second one we get $x^2-2x+1=0$ , or $x=1$ .

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