Question

Решите уравнение:
x^4+8x-7=0

Answer 1

$x^4+8x-7=0\\x^4+1+8x-8=0\\(x^2+1)^2-2x^2+8x-8=0\\(x^2+1)^2-2(x^2-8x+4)=0\\(x^2+1)^2-2(x-2)^2=0\\(x^2+1)^2-(\sqrt{2}x-2\sqrt{2})^2=0\\(x^2+1+\sqrt{2}x-2\sqrt{2})(x^2+1-\sqrt{2}x+2\sqrt{2})=0\\1)\\x^2+\sqrt{2}x+(1-2\sqrt{2})=0\\D=2-4(1-2\sqrt{2})=2-4+8\sqrt{2}=-2+\sqrt{8}\\x_1=\frac{-\sqrt{2}+\sqrt{-2+\sqrt{8} } }{2} ; x_2=\frac{-\sqrt{2}-\sqrt{-2+\sqrt{8} } }{2} \\2)\\x^2-\sqrt{2}x+(1+2\sqrt{2})=0\\D=2-4(1+2\sqrt{2})=2-4-8\sqrt{2}=-2-8\sqrt{2}<0$

x ∈ ∅

Ответ: $x_1=\frac{-\sqrt{2}+\sqrt{-2+\sqrt{8} } }{2} ; x_2=\frac{-\sqrt{2}-\sqrt{-2+\sqrt{8} } }{2}$