Найдите корни уравнения: 3(x^2+1/x^2)+7(x+1/x)=4.

\[3 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) + 7 \cdot \left( x + \frac{1}{x} \right) = 4\]

\[3 \cdot \left( \left( x + \frac{1}{x} \right)^{2} - 2 \right) + 7 \cdot \left( x + \frac{1}{x} \right) =\]

\[= 4\]

\[Пусть\ \left( x + \frac{1}{x} \right) = a:\]

\[3 \cdot \left( a^{2} - 2 \right) + 7a - 4 = 0\]

\[3a^{2} - 6 + 7a - 4 = 0\]

\[3a^{2} + 7a - 10 = 0\]

\[D = 49 + 120 = 169\]

\[a_{1} = \frac{- 7 + 13}{6} = 1;\ \ \]

\[a_{2} = \frac{- 7 - 13}{6} = - \frac{20}{6} = - \frac{10}{3}.\]

\[Подставим:\]

\[1)\ x + \frac{1}{x} = 1\ \ \ \ \ \ \ \ | \cdot x\]

\[x^{2} + 1 - x = 0\]

\[x^{2} - x + 1 = 0\]

\[D = 1 - 4 = - 3 < 0\]

\[нет\ корней.\]

\[2)\ x + \frac{1}{x} = - \frac{10}{3}\ \ \ \ \ \ \ | \cdot 3x\]

\[3x^{2} + 3 + 10x = 0\]

\[3x^{2} + 10x + 3 = 0\]

\[D = 25 - 9 = 16\]

\[x_{1} = \frac{- 5 + 4}{3} = - \frac{1}{3};\ \ \ \]

\[x_{2} = \frac{- 5 - 4}{3} = - 3.\]

\[Ответ:x = - \frac{1}{3};\ \ \ x = - 3.\]