Question

(4+2х)(12-3х)> 0
помогите

Answer 1

$(4+2x)(12-3x)\ \textgreater \ 0 \\ 48-12x+24x-6x^2\ \textgreater \ 0 \\ (-6x^2+12x+48\ \textgreater \ 0):(-6) \\ x^2-2x-8\ \textless \ 0 \\ x^2-2x-8=0 \\ Po Teoreme obratnoy k teoreme Vieta: \\ \left \{ {{x_1+x_2=2} \atop {x_1*x_2=-8}} \right. \\ \left \{ {{x_1=4} \atop {x_2=-2}} \right. \\ x_1\ \textless \ 4 \\ x_2\ \textless \ -2$