Bonsoir,
[tex]a) \ f(x) = (x-7)(2x+4) \iff 2x^2 + 4x -14x -28 \iff \boxed{2x^2 -10x -28}[/tex]
[tex]b)[/tex]
[tex]\textnormal{M\'ethode 1 :}[/tex]
[tex]\Delta = (-10)^2 -4 \times 2 \times (-28) = 324[/tex]
[tex]x_1 = \frac{-b-\sqrt{\Delta} }{2a} = \frac{10-\sqrt{324} }{4} = -2[/tex]
[tex]x_1 = \frac{-b+\sqrt{\Delta} }{2a} = \frac{10+\sqrt{324} }{4} = 7[/tex]
[tex]x_1 + x_2 = -2 + 7 = 5[/tex]
[tex]x_1 \times x_2 =-2 \times 7 = -14[/tex]
[tex]\textnormal{M\'ethode 2 :}[/tex]
[tex]x_1 + x_2 = -\frac{b}{a} = -(\frac{-10}{2}) =5[/tex]
[tex]x_1 \times x_2 = \frac{c}{a} = -\frac{28}{2} = -14[/tex]
[tex]\textnormal{On peut donc en conclure que Wesley a bien raison.}[/tex]
