Для каждого значения а решите уравнение: x^2+(1-3a)x+2a^2-2=0.

\[x² + (1 - 3a)x + 2a² - 2 = 0\]

\[= 1 - 6a + 9a^{2} - 8a^{2} + 8 =\]

\[= a^{2} - 6a + 9 = (a - 3)^{2}\]

\[x_{1,2} = \frac{- (1 - 3a) \pm \sqrt{(a - 3)^{2}}}{2 \cdot 1} =\]

\[= \frac{3a - 1 \pm |a - 3|}{2}\]

\[1)\ \ a = 3:\]

\[x = \frac{3 \cdot 3 - 1 \pm |3 - 3|}{2} =\]

\[= \frac{9 - 1 \pm 0}{2} = \frac{8}{2} = 4.\]

\[2)\ a > 3:\]

\[x_{1} = \frac{3a - 1 + a - 3}{2} = \frac{4a - 4}{2} =\]

\[= 2a - 2;\]

\[x_{2} = \frac{3a - 1 - a + 3}{2} = \frac{2a + 2}{2} =\]

\[= a + 1.\]

\[Ответ:a = 3 \Longrightarrow \ x = 4;\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ a > 3 \Longrightarrow x_{1} = 2a - 2;\ \ \]

\[x_{2} = a + 1.\]