Помогите решить уравнение:. Created by alkolokoltseva. matematika-ru

Помогите решить уравнение:...

julyap
$x^{2} +2(1+ \sqrt{8} )x+8 \sqrt{2} =0 \\ x^{2} +2x+2x \sqrt{8} +8 \sqrt{2} =0\\ x^{2} +2x+2x* \sqrt{4*2} +8 \sqrt{2} =0\\ x^{2} +2x+4x \sqrt{2} +8 \sqrt{2} =0\\x(x+2)+4 \sqrt{2} (x+2)=0$

$\\(x+2)(x+4 \sqrt{2} )=0\\x+2=0\\ x=-2\\\\x+4 \sqrt{2} =0\\x=-4 \sqrt{2}$

1 votes Thanks 1

wwwElIzAvEtAwww X²+2(1+√8)x+8√2=0
Можно решать так:
D=(2(1+√8))²-4•1•8√2=4+8√8+32-32√2=4-16√2+32=(2-4√2)²
√D=|2-4√2|=4√2-2 (2-4√2<0)
x=(-2(1+√8)±(4√2-2))/2=-1-√8±(2√2-1)
x₁=-2;x₂=-4√2
Ответ: -4√2;-2.

0 votes Thanks 1

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