Question

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Answer 1

$Evident~x \neq 0. \\ \\ Notez ~\frac{x^2+1}{x}=a. ~Ecuatia~devine:~a+ \frac{1}{a}=2,9 \Leftrightarrow \\ \\ \Leftrightarrow a^2+1=2,9a \Leftrightarrow a^2-2,9a+1=0. \\ \\ \Delta =(-2,9)^2 -4=4,41. \\ \\ a_{1}= \frac{2,9 + \sqrt{ \Delta}}{2}= \frac{2,9 +2,1}{2}= \frac{5}{2}. \\ \\ a_2= \frac{2,9 - \sqrt{ \Delta}}{2}= \frac{2,9-2,1}{2}=0,4= \frac{2}{5}.$

$\boxed{1.}~ a= \frac{5}{2} \Rightarrow \frac{x^2+1}{x}= \frac{5}{2} \Rightarrow 2x^2+2=5x. \\ \\ 2x^2-5x+2=0. \\ \\ \Delta=(-5)^2-4 \cdot 2 \cdot 2=25-16=9. \\ \\ x_{1}= \frac{5 + \sqrt{ \Delta}}{4}= \frac{5+3}{4}=2. \\ \\ x_2= \frac{5- \sqrt{ \Delta}}{4}= \frac{5-3}{4}= \frac{1}{2}.$

$\boxed{2.}~a= \frac{2}{5} \Rightarrow \frac{x^2+1}{x} = \frac{2}{5} \Rightarrow 5x^2+5=2x. \\ \\ 5x^2-2x+5=0. \\ \\ \Delta=(-2)^2-4 \cdot 5 \cdot 5= -96 \ \textless \ 0 \Rightarrow ~ecuatia~nu~are~solutii~reale. \\ \\ In~concluzie,~solutiile~ecuatiei~sunt~ x \in \Big \{ \frac{1}{2}~;~2 \Big\}.$