Вопрос:

Для каждого значения а решите уравнение: 3(a-2)x^2+(a-5)x-1=0.

Ответ:

\[3 \cdot (a - 2)x² + (a - 5)x - 1 = 0\]

\[= a^{2} - 10a + 25 + 12a - 24 =\]

\[= a^{2} + 2a + 1 = (a + 1)^{2}\]

\[При\ D = 0:\]

\[(a + 1)^{2} = 0\]

\[a + 1 = 0\]

\[a = - 1.\]

\[1)\ D = 0;\ a = - 1:\]

\[x = \frac{5 - ( - 1) \pm | - 1 + 1|}{6 \cdot ( - 1 - 2)} =\]

\[= \frac{6}{- 18} = - \frac{1}{3}.\]

\[2)\ D > 0;a > - 1:\]

\[x_{1} = \frac{5 - a + a + 1}{6 \cdot (a - 2)} =\]

\[= \frac{6}{6 \cdot (a - 2)} = \frac{1}{a - 2};\ \]

\[x_{2} = \frac{5 - a - a - 1}{6 \cdot (a - 2)} =\]

\[3)\ D < 0 \Longrightarrow корней\ нет.\]

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

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

\[или\ x_{2} = - \frac{1}{3}.\]


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