Question

Решите уравнение $x^{2} +4x+2=0$

Answer 1

Ответ:

$x_{1}=-2+\sqrt{2} ; x_{2}=-2-\sqrt{2}$

Объяснение:

$x^2+4x+2=0\\D=b^2-4ac=16-8=8\\x_{1}=\frac{-b+\sqrt{D} }{2a}=\frac{-4+\sqrt{8} }{2} =\frac{-4+2\sqrt{2} }{2}=\frac{2(-2+\sqrt{2}) }{2}= -2+\sqrt{2} \\x_{2}=\frac{-b-\sqrt{D} }{2a}=\frac{-4-\sqrt{8} }{2} =\frac{-4-2\sqrt{2} }{2}=\frac{2(-2-\sqrt{2}) }{2}= -2-\sqrt{2}$

Answer 2

x² + 4x + 2 = 0        D = b²-4ac = 16-8 = 8

x₁ = (-b+√D)/2a = (-4+2√2)/2 = -2+√2    

x₂ = (-b-√D)/2a = (-4-2√2)/2 = -2-√2

Ответ: {-2-√2; -2+√2}