Question

X la a doua+6x+8=0 rezolvati!!!

Answer 1

[tex] x^{2} +6x+8=0 \\ x^{2} +4x+2x+8=0 \\ x(x+4)+2(x+4)=0 \\ (x+4)(x+2)=0=\ \textgreater \ \\ x_{1} =-4 \\ x_{2}=-2[/tex]

Answer 2

$x^{2} +6x+8=0 \\ \Delta= b^{2}-4ac=6 ^{2} -4*1*8= 4 \\ \\ x_{1,2}= \frac{-b \pm \sqrt{ \Delta} }{2a}= \frac{-6 \pm 2}{2} = \left \{ {{-4} \atop {-2}} \right. . \\ \\ ------sau------ \\ x^{2} +6x+8=0 \Leftrightarrow x^{2} +6x+9-1=0 \Leftrightarrow (x+3)^{2}-1=0 \Leftrightarrow \\ \Leftrightarrow (x+4)(x+2)=0 \Rightarrow \boxed{\boxed{x=-4}~sau~\boxed{x=-2}}. \\ \\ Solutie:~\boxed{x \in\{-4;-2\}}.$