$x^4 - 29x^2 + 100 = 0$
$x^2 = y$, y ≥ 0
$y^2 - 29y + 100 = 0$
$D = b^2 - 4ac = (-29)^2 - 4 * 1 * 100 = 841 - 400 = 441 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{29 + \sqrt{441}}{2 * 1} = \frac{29 + 21}{2} = \frac{50}{2} = 25$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{29 - \sqrt{441}}{2 * 1} = \frac{29 - 21}{2} = \frac{8}{2} = 4$
$x^2 = 25$
x = ±5
или
$x^2 = 4$
x = ±2
Ответ: −5; −2; 2; 5.

$x^4 - 9x^2 + 20 = 0$
$x^2 = y$, y ≥ 0
$y^2 - 9y + 20 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 1 * 20 = 81 - 80 = 1 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{1}}{2 * 1} = \frac{9 + 1}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{1}}{2 * 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$
$x^2 = 5$
$x = ±\sqrt{5}$
или
$x^2 = 4$
x = ±2
Ответ: $-\sqrt{5}; -2; 2; \sqrt{5}$.

$x^4 - 2x^2 - 24 = 0$
$x^2 = y$, y ≥ 0
$y^2 - 2y - 24 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-24) = 4 + 96 = 100 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{100}}{2 * 1} = \frac{2 + 10}{2} = \frac{12}{2} = 6$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{100}}{2 * 1} = \frac{2 - 10}{2} = \frac{-8}{2} = -4$ − не удовлетворяет условию, так как y ≥ 0.
$x^2 = 6$
$x = ±\sqrt{6}$
Ответ: $-\sqrt{6}; \sqrt{6}$.

$x^4 + 3x^2 - 70 = 0$
$x^2 = y$, y ≥ 0
$y^2 + 3y - 70 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * (-70) = 9 + 280 = 289 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{289}}{2 * 1} = \frac{-3 + 17}{2} = \frac{14}{2} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{289}}{2 * 1} = \frac{-3 - 17}{2} = \frac{-20}{2} = -10$ − не удовлетворяет условию, так как y ≥ 0.
$x^2 = 7$
$x = ±\sqrt{7}$
Ответ: $-\sqrt{7}; \sqrt{7}$.

$9x^4 - 10x^2 + 1 = 0$
$x^2 = y$, y ≥ 0
$9y^2 - 10y + 1 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 9 * 1 = 100 - 36 = 64 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{64}}{2 * 9} = \frac{10 + 8}{18} = \frac{18}{18} = 1$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{64}}{2 * 9} = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}$
$x^2 = 1$
x = ±1
или
$x^2 = \frac{1}{9}$
$x = ±\sqrt{\frac{1}{3}}$
Ответ: $-1; -\sqrt{\frac{1}{3}}; \sqrt{\frac{1}{3}}; 1$.

$2x^4 - 5x^2 + 2 = 0$
$x^2 = y$, y ≥ 0
$2y^2 - 5y + 2 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 2 * 2 = 25 - 16 = 9 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{9}}{2 * 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{9}}{2 * 2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}$
$x^2 = 2$
$x = ±\sqrt{2}$
или
$x^2 = \frac{1}{2}$
$x = ±\sqrt{\frac{1}{2}}$
Ответ: $-\sqrt{2}; -\sqrt{\frac{1}{2}}; \sqrt{\frac{1}{2}}; \sqrt{2}$.