Question

ajutatima si pe mine va rog​

Answer 1

Explicație pas cu pas:

1)

$2 {x}^{2} + 6x + 4 = 0 \\ 2({x}^{2} + 3x + 2) = 0 \\ 2(x + 1)(x + 2) = 0 \\ x + 1 = 0 \implies x = - 1 \\ x + 2 = 0 \implies x = - 2$

2)

$3 {x}^{2} - 6x + 3 = 0 \\ 3 ({x}^{2} - 2x + 1) = 0 \\ 3 {(x - 1)}^{2} = 0 \\ x - 1 = 0 \implies x = 1$

3)

${x}^{2} - x + 1 = 0 \\\Delta = 1 - 4 = - 3 < 0 \\ \implies x \not\in \mathbb{R}$