Question

ТЕРМІНОВО! 10 клас
Функцію задано формулою f (x) = –3x2 + 2x. Знайдіть значення аргументу, при якому значення функції f дорівнює: 0; –1; –56.

Answer 1

$f(x) = - 3 {x}^{2} + 2x$

$1) \: f(x) = 0 \\ - 3 {x}^{2} + 2x = 0 \\ 3 {x}^{2} - 2x = 0 \: \: | \div 3 \\ {x}^{2} - \frac{2}{3} x = 0 \\ x(x - \frac{2}{3} ) = 0 \\ x_{1} =0 \\ x_{2} = \frac{2}{3}$

$2) \: f(x) = - 1 \\ - 3 {x}^{2} + 2x = - 1 \\ - 3 {x}^{2} + 2x + 1 = 0 \\ 3 {x}^{2} - 2x - 1 = 0 \\ a = 3 \\ b = - 2 \\ c = - 1 \\ D = b {}^{2} - 4ac = ( - 2) {}^{2} - 4 \times 3 \times ( - 1) =4 + 12 = 16 \\ x_{1} = \frac{2 + 4}{6} = \frac{6}{6} = 1\\ x_{2} = \frac{2 - 4}{6} = \frac{ - 2}{6} = - \frac{1}{3}$

$3) \: f(x) = - 56 \\ - 3 {x}^{2} + 2x = - 56 \\ - 3 {x}^{2} + 2x + 56 = 0 \\ 3 {x}^{2} - 2x - 56 = 0 \\ a =3 \\ b = - 2\\ c = - 56\\ D = b {}^{2} - 4ac = ( - 2) {}^{2} - 4 \times 3 \times ( - 56) = \\ = 4 + 672 = 676 \: \: ( \sqrt{D} = 26) \\ x_{1} = \frac{2 + 26}{6} = \frac{28}{6} = \frac{14}{3} = 4 \frac{2}{3} \\ x_{2} = \frac{2 - 26}{6} = \frac{ - 24}{6} = - 4$