$x - 6\sqrt{x} + 8 = 0$
$(\sqrt{x})^2 - 6\sqrt{x} + 8 = 0$
$\sqrt{x} = y$, y ≥ 0
$y^2 - 6y + 8 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * 8 = 36 - 32 = 4 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{4}}{2 * 1} = \frac{6 + 2}{2} = \frac{8}{2} = 4$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{4}}{2 * 1} = \frac{6 - 2}{2} = \frac{4}{2} = 2$
$\sqrt{x} = 4$
$(\sqrt{x})^2 = 4^2$
x = 16
или
$\sqrt{x} = 2$
$(\sqrt{x})^2 = 2^2$
x = 4
Ответ: 4 и 16

$x - 5\sqrt{x} - 50 = 0$
$(\sqrt{x})^2 - 5\sqrt{x} - 50 = 0$
$\sqrt{x} = y$, y ≥ 0
$y^2 - 5y - 50 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * (-50) = 25 + 200 = 225 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{225}}{2 * 1} = \frac{5 + 15}{2} = \frac{20}{2} = 10$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{225}}{2 * 1} = \frac{5 - 15}{2} = \frac{-10}{2} = -5$ − не удовлетворяет условию, так как y ≥ 0.
$\sqrt{x} = 10$
$(\sqrt{x})^2 = 10^2$
x = 100
Ответ: 100

$2x - 3\sqrt{x} + 1 = 0$
$2(\sqrt{x})^2 - 3\sqrt{x} + 1 = 0$
$\sqrt{x} = y$, y ≥ 0
$2y^2 - 3y + 1 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 2 * 1 = 9 - 8 = 1 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 2} = \frac{3 + 1}{4} = \frac{4}{4} = 1$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 2} = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$
$\sqrt{x} = 1$
$(\sqrt{x})^2 = 1^2$
x = 1
или
$\sqrt{x} = \frac{1}{2}$
$(\sqrt{x})^2 = (\frac{1}{2})^2$
$x = \frac{1}{4}$
Ответ: $\frac{1}{4}$ и 1