Решите уравнение и выполните проверку по теореме, обратной теореме Виета:
а) $x^2 - 5\sqrt{2}x + 12 = 0$;
б) $x^2 + 2\sqrt{3}x - 72 = 0$;
в) $y^2 - 6y + 7 = 0$;
г) $p^2 - 10p + 7 = 0$.
$x^2 - 5\sqrt{2}x + 12 = 0$
D = 25 * 2 − 4 * 12 = 2
$x = \frac{5\sqrt{2} ± \sqrt{2}}{2}$
$x_1 = \frac{5\sqrt{2} - \sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$
$x_2 = \frac{5\sqrt{2} + \sqrt{2}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$
Проверка:
$x_1 + x_2 = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$
$x_1x_2 = 2\sqrt{2} * 3\sqrt{2} = 6 * 2 = 12$
$x^2 + 2\sqrt{3}x - 72 = 0$
D = 3 + 72 = 75
$x = -\sqrt{3} ± \sqrt{75} = -\sqrt{3} ± 5\sqrt{3}$
$x_1 = -\sqrt{3} - 5\sqrt{3} = -6\sqrt{3}$
$x_2 = -\sqrt{3} + 5\sqrt{3} = 4\sqrt{3}$
Проверка:
$x_1 + x_2 = -6\sqrt{3} + 4\sqrt{3} = -2\sqrt{3}$
$x_1x_2 = -6\sqrt{3} * 4\sqrt{3} = -24 * 3 = -72$
$y^2 - 6y + 7 = 0$
$D = 3^2 - 7 = 2$
$y = 3 ± \sqrt{2}$
$y_1 = 3 - \sqrt{2}$
$y_2 = 3 + \sqrt{2}$
Проверка:
$y_1 + y_2 = 3 - \sqrt{2} + 3 + \sqrt{2} = 6$
$y_1y_2 = (3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$
$p^2 - 10p + 7 = 0$
$D = 5^2 - 7 = 18$
$p = 5 ± \sqrt{18} = 5 ± 3\sqrt{2}$
$p_1 = 5 - 3\sqrt{2}$
$p_2 = 5 + 3\sqrt{2}$
Проверка:
$p_1 + p_2 = 5 - 3\sqrt{2} + 5 + 3\sqrt{2} = 10$
$p_1p_2 = (5 - 3\sqrt{2})(5 + 3\sqrt{2}) = 25 - 9 * 2 = 25 - 18 = 7$
Пожауйста, оцените решение