Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 111

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\[1)\ x^{2} - 12x + 35 =\]

\[= (x - 5)(x - 7)\]

\[D_{1} = 36 - 35 = 1\]

\[x_{1} = 6 + 1 = 7;\ \ \]

\[x_{2} = 6 - 1 = 5.\]

\[2)\ x^{2} + 9x + 20 =\]

\[= (x + 5)(x + 4)\]

\[x_{1} + x_{2} = - 9;\ \ x_{1} \cdot x_{2} = 20\]

\[x_{1} = - 5;\ \ x_{2} = - 4.\]

\[3)\ 5x^{2} + 9x - 2 =\]

\[= 5 \cdot \left( x - \frac{1}{5} \right)(x + 2) =\]

\[= (5x - 1)(x + 2)\]

\[D = 81 + 40 = 121\]

\[x_{1} = \frac{- 9 + 11}{10} = \frac{2}{10} = \frac{1}{5};\ \]

\[x_{2} = \frac{- 9 - 11}{10} = - \frac{20}{10} = - 2.\]

\[4)\ 4x^{2} - x - 3 =\]

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

\[= (4x + 3)(x - 1)\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{8} = 1;\ \ \ \]

\[x_{2} = \frac{1 - 7}{8} = - \frac{6}{8} = - \frac{3}{4}.\]

\[5) - 2x^{2} + 5x - 2 =\]

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

\[= (2x - 1)(2 - x)\]

\[2x^{2} - 5x + 2 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 + 3}{4} = 2;\ \ \]

\[\ x_{2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}.\]

\[6)\ \frac{2}{3}x^{2} + 2x - 12 =\]

\[= \frac{2}{3} \cdot (x + 6)(x - 3)\]

\[\frac{2}{3}x^{2} + 2x - 12 = 0\ \ \ \ | \cdot \frac{3}{2}\]

\[x^{2} + 3x - 18 = 0\]

\[x_{1} + x_{2} = - 3;\ \ \ x_{1} \cdot x_{2} = - 18\]

\[x_{1} = - 6;\ \ x_{2} = 3.\]