Заполните таблицу, где a, b и c − коэффициенты квадратного уравнения $ax^2 + bx + x = 0$, а $x_1$ и $x_2$ − его корни.
| Уравнение | $-\frac{b}{a}$ | $\frac{c}{a}$ | $x_1$ | $x_2$ | $x_1+ x_2$ | $x_1x_2$ |
|---|---|---|---|---|---|---|
| $7x^2 - 8x + 1=0$ | ||||||
| $6x^2 + 13x - 15=0$ |
$7x^2 - 8x + 1=0$
$-\frac{b}{a} = -\frac{-8}{7} = \frac{8}{7} = 1\frac{1}{7}$
$\frac{c}{a} = \frac{1}{7}$
$D = b^2 - 4ac = (-8)^2 - 4 * 7 * 1 = 64 - 28 = 36$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{8 + \sqrt{36}}{2 * 7} = \frac{8 + 6}{14} = \frac{14}{14} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{8 - \sqrt{36}}{2 * 7} = \frac{8 - 6}{14} = \frac{2}{14} = \frac{1}{7}$
$x_1 + x_2 = 1 + \frac{1}{7} = 1\frac{1}{7}$
$x_1 * x_2 = 1 * \frac{1}{7} = \frac{1}{7}$
$6x^2 + 13x - 15=0$
$-\frac{b}{a} = -\frac{13}{6} = -2\frac{1}{6}$
$\frac{c}{a} = \frac{-15}{6} = -\frac{5}{2} = -2\frac{1}{2}$
$D = b^2 - 4ac = 13^2 - 4 * 6 * (-15) = 169 + 360 = 529$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-13 + \sqrt{529}}{2 * 6} = \frac{-13 + 23}{12} = \frac{10}{12} = \frac{5}{6}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-13 - \sqrt{529}}{2 * 6} = \frac{-13 - 23}{12} = \frac{-36}{12} = -3$
$x_1 + x_2 = \frac{5}{6} + (-3) = \frac{5}{6} - 2\frac{6}{6} = -2\frac{1}{6}$
$x_1 * x_2 = \frac{5}{6} * (-3) = -\frac{5}{2} = -2\frac{1}{2}$
Ответ:
| Уравнение | $-\frac{b}{a}$ | $\frac{c}{a}$ | $x_1$ | $x_2$ | $x_1+ x_2$ | $x_1x_2$ |
|---|---|---|---|---|---|---|
| $7x^2 - 8x + 1=0$ | $1\frac{1}{7}$ | $\frac{1}{7}$ | $1$ | $\frac{1}{7}$ | $1\frac{1}{7}$ | $\frac{1}{7}$ |
| $6x^2 + 13x - 15=0$ | $-2\frac{1}{6}$ | $-2\frac{1}{2}$ | $\frac{5}{6}$ | $-3$ | $-2\frac{1}{6}$ | $-2\frac{1}{2}$ |
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