Question

Решите пж пж пж срочно надо
8x^2-x-9=0
x^4-16x^2+28=0 ^-степень 2 решить биквадратное
пж пж пж пж пж

Answer 1

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$8 {x}^{2} - x - 9 = 0 \ d = 1 - 4 times 8 times ( - 9) = 1 + 288 = 289 \ x1 = frac{1 + sqrt{d} }{2 times 8} = frac{1 + sqrt{289} }{16} = frac{1 + 17}{16} = frac{18}{16} = frac{9}{8} = 1 frac{1}{8} = 1.125 \ x2 = frac{1 - sqrt{289} }{16} = frac{1 - 17}{16} = frac{ - 16}{16} = - 1$
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${x}^{4} - 16 {x}^{2} + 28 = 0 \ {x}^{2} = y \ {y}^{2} - 16y + 28 = 0 \ d = {16}^{2} - 4 times 28 = 256 - 112 = 144 \ y1 = frac{16 + sqrt{144} }{2} = frac{16 + 12}{2} = frac{28}{2} = 14 \ y2 = frac{16 - sqrt{144} }{2} = frac{16 - 12}{2} = frac{4}{2} = 2 \ - - - - - - - - - - \ {x}^{2} = y1 \ {x}^{2} = 14 \ x1 = sqrt{14} \ x3 = - sqrt{14} \ - - - - - - - - - - \ {x}^{2} = y2 \ {x}^{2} = 2 \ x 2= sqrt{2} \x4 = - sqrt{2}$