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

$x^4 - 5x^2 + 6 = 0$
$x^2 = y$, y ≥ 0
$y^2 - 5y + 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * 6 = 25 - 24 = 1 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{1}}{2 * 1} = \frac{5 + 1}{2} = \frac{6}{2} = 3$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{1}}{2 * 1} = \frac{5 - 1}{2} = \frac{4}{2} = 2$
$x^2 = 3$
$x = ±\sqrt{3}$
или
$x^2 = 2$
$x = ±\sqrt{2}$
Ответ: $-\sqrt{3}; -\sqrt{2}; \sqrt{2}; \sqrt{3}$.

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

$x^4 + 14x^2 - 32 = 0$
$x^2 = y$, y ≥ 0
$y^2 + 14y - 32 = 0$
$D = b^2 - 4ac = 14^2 - 4 * 1 * (-32) = 196 + 128 = 324 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-14 + \sqrt{324}}{2 * 1} = \frac{-14 + 18}{2} = \frac{4}{2} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-14 - \sqrt{324}}{2 * 1} = \frac{-14 - 18}{2} = \frac{-32}{2} = -16$ − не удовлетворяет условию, так как y ≥ 0.
$x^2 = 2$
$x = ±\sqrt{2}$
Ответ: $-\sqrt{2}; \sqrt{2}$.

$4x^4 - 9x^2 + 2 = 0$
$x^2 = y$, y ≥ 0
$4y^2 - 9y + 2 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 4 * 2 = 81 - 32 = 49 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{49}}{2 * 4} = \frac{9 + 7}{8} = \frac{16}{8} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{49}}{2 * 4} = \frac{9 - 7}{8} = \frac{2}{8} = \frac{1}{4}$
$x^2 = 2$
$x = ±\sqrt{2}$
или
$x^2 = \frac{1}{4}$
$x = ±\frac{1}{2}$
Ответ: $-\sqrt{2}; -\frac{1}{2}; \frac{1}{2}; \sqrt{2}$.

$3x^4 + 8x^2 - 3 = 0$
$x^2 = y$, y ≥ 0
$3y^2 + 8y - 3 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 3 * (-3) = 64 + 36 = 100 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{100}}{2 * 3} = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{100}}{2 * 3} = \frac{-8 - 10}{6} = \frac{-18}{6} = -3$ − не удовлетворяет условию, так как y ≥ 0.
$x^2 = \frac{1}{3}$
$x = ±\sqrt{\frac{1}{3}}$
Ответ: $-\sqrt{\frac{1}{3}}; \sqrt{\frac{1}{3}}$.