Решить уравнение [tex]x^4+5x^3+4x^2-15x-21=0[/tex]. Created by yugolovin. algebra-ru

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vimmortal2

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Ответ:

-√3; √3

Объяснение:

x⁴ + 5x³ + 4x² - 15x - 21 = 0

x⁴ + 5x³ + 7x² - 3x² - 15x - 21 = 0

x²(x² - 3) + 5x(x² - 3) + 7(x² - 3) = 0

(x² + 5x + 7)(x² - 3) = 0

Продолжение на фото

2 votes Thanks 4

Artsounds

Ответ:

$x {}^{4} + 5x {}^{3} + 4x {}^{2} - 15x - 21 = 0 \\ x {}^{4} + 5x {}^{3} - 3x {}^{2} + 7x {}^{2} - 15x - 21 = 0 \\ x {}^{2} \times ( {x}^{2} - 3) + 5x \times ( {x}^{2} - 3) + 7 \times ( {x}^{2} - 3) = 0 \\ (x {}^{2} - 3) \times (x {}^{2} + 5x + 7) = 0 \\ x {}^{2} - 3 = 0 \: \: \: \: \: \: \: \: \: \: {x}^{2} + 5x + 7 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x {}^{2} = 0 + 3 \: \: \: \: \: \: \: \: \boxed{x = \frac{ - b \pm \sqrt{b {}^{2} - 4ac} }{2a} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x {}^{2} = 3 \ \: \: \: \: \: \: \: \: x = \frac{ - 5 \pm \sqrt{5 {}^{2} - 4 \times 1 \times 7} }{2 \times 1} \\ \\ x = \pm \sqrt{3 } \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 5 \pm \sqrt{ - 3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \\x \notin \huge \: r \\ \\ \otvet : \boxed{x = \pm \sqrt{3} }$

2 votes Thanks 0

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