Найдите корни уравнения и выполните проверку по теореме, обратной теореме Виета:
а) $x^2 - 15x - 16 = 0$;
б) $x^2 - 6x - 11 = 0$;
в) $12x^2 - 4x - 1 = 0$;
г) $x^2 - 6 = 0$;
д) $5x^2 - 18x = 0$;
е) $2x^2 - 41 = 0$.
$x^2 - 15x - 16 = 0$
$D = 15^2 + 4 * 16 = 225 + 64 = 289$
$x = \frac{15 ± \sqrt{289}}{2}$
$x_1 = \frac{15 - 17}{2} = \frac{-2}{2} = -1$
$x_2 = \frac{15 + 17}{2} = \frac{32}{2} = 16$
Проверка:
$x_1 + x_2 = -1 + 16 = 15$;
$x_1x_2 = -1 * 16 = -16$.
$x^2 - 6x - 11 = 0$
$D = 3^2 + 11 = 9 + 11 = 20$
$x = 3 ± \sqrt{20} = 3 ± 2\sqrt{5}$
$x_1 = 3 - 2\sqrt{5}$
$x_2 = 3 + 2\sqrt{5}$
Проверка:
$x_1 + x_2 = 3 - 2\sqrt{5} + 3 + 2\sqrt{5} = 6$;
$x_1x_2 = (3 - 2\sqrt{5})(3 + 2\sqrt{5}) = 9 - 20 = -11$.
$12x^2 - 4x - 1 = 0$
$D = 2^2 + 12 * 1 = 4 + 12 = 16$
$x = \frac{2 ± \sqrt{16}}{12}$
$x_1 = \frac{2 - 4}{12} = \frac{-2}{12} = -\frac{1}{6}$
$x_2 = \frac{2 + 4}{12} = \frac{6}{12} = \frac{1}{2}$
Проверка:
$12(x_1 + x_2) = 12(-\frac{1}{6} + \frac{1}{2}) = 12(-\frac{1}{6} + \frac{3}{6}) = 12 * \frac{2}{6} = 4$;
$12_1x_2 = 12 * (-\frac{1}{6}) * \frac{1}{2} = -1$.
$x^2 - 6 = 0$
$x = ±\sqrt{6}$
$x_1 = -\sqrt{6}$
$x_2 = \sqrt{6}$
Проверка:
$x_1 + x_2 = -\sqrt{6} + \sqrt{6} = 0$;
$x_1x_2 = -\sqrt{6} * \sqrt{6} = -6$.
$5x^2 - 18x = 0$
x(5x − 18) = 0
$x_1 = 0$
5x = 18
$x_2 = 3,6$
Проверка:
$5(x_1 + x_2) = 5(0 + 3,6) = 18$;
$5x_1x_2 = 5 * 0 * 3,6 = 0$.
$2x^2 - 41 = 0$
$2x^2 = 41$
$x^2 = 20,5$
$x = ±\sqrt{20,5}$
$x_1 = -\sqrt{20,5}$
$x_2 = \sqrt{20,5}$
Проверка:
$2(x_1 + x_2) = 2(-\sqrt{20,5} + \sqrt{20,5}) = 0$;
$2x_1x_2 = 2 * (-\sqrt{20,5}) * \sqrt{20,5} = 2 * (-20,5) = -41$.
Пожауйста, оцените решение