Question

2x^2+3x+1=0 решите квадратное уравнение

Answer 1

Ответ:

2x²+3х+1=0

D=b²-4ac=3²-4*2*1=9-8=1

$x_{1,2}=\frac{-bб\sqrt{D} }{2a}$

$x_{1}=\frac{-3+\sqrt{1} }{2*2}=\frac{-3+1}{4}=\frac{-2}{4}=-\frac{1}{2}=-0,5$

$x_{2}=\frac{-3-\sqrt{1} }{2*2}=\frac{-3-1}{4}=\frac{-4}{4}=-1$

Ответ: х₁= -0,5, х₂= -1

Answer 2

$2x^{2} + 3x + 1 = 0 \\ 2x \times (x + 1) + x + 1 = 0 \\ (x + 1) \times (2x + 1) = 0 \\ x + 1 = 0 \\ 2x + 1 = 0 \\ x = - 1 \\ x = - \frac{1}{2} \\ x1 = - 1 \\ x2 = - \frac{1}{2}$