ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 229

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\[\textbf{а)}\ 3x² - 25x - 28 = 0\]

\[D = 25 + 4 \cdot 3 \cdot 28 = 961\]

\[x_{1,2} = \frac{25 \pm 31}{6},\ \ x_{1} = \frac{28}{3},\ \ \]

\[x_{2} = - 1;\]

\[\Longrightarrow 3x^{2} - 25x - 28 =\]

\[= 3 \cdot \left( x - \frac{28}{3} \right)(x + 1) =\]

\[= (3x - 28)(x + 1);\]

\[\textbf{б)}\ 2x² + 13x - 7 = 0\]

\[D = \ 169 + 4 \cdot 2 \cdot 7 = 225\]

\[x_{1,2} = \frac{- 13 \pm 15}{4},\ \ x_{1} = \frac{1}{2},\ \ \]

\[x_{2} = - 7;\]

\[\Longrightarrow 2x^{2} + 13x - 7 =\]

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

\[= (2x - 1)(x + 7).\]