Bonjour,
[tex](2x+1)^2 - (x-3)(2x+1 ) = 0 \Leftrightarrow \\
(2x+1)[(2x+1) - (x-3)] = 0 \Leftrightarrow \\
(2x+1)(x+4) = 0 \Leftrightarrow\\
\left \{ {{2x+1 = 0 \ ou} \atop {x+4 = 0}} \right.
\Leftrightarrow
\left \{ {{x = -\frac{1}{2} \ ou} \atop {x = -4}} \right. \\
[/tex]
[tex](2x+1)^2 - (x-3)(2x+1 ) = 4
\Leftrightarrow \\
4x^2 + 4x + 1 - 2x^2 - x + 6x + 3 = 4
\Leftrightarrow \\
2x^2 + 10x + 4 = 4
\Leftrightarrow \\
2x^2 + 10x = 0
\Leftrightarrow\\
2x(x+5) = 0
\Leftrightarrow\\
\left \{ {{x= 0 \ ou} \atop {x+5 = 0}} \right.
\Leftrightarrow
\left \{ {{x = 0} \ ou} \atop {x = -5}} \right. \\[/tex]
