Question

Va rogggg ajutati-ma!!!!!!!!
Determinati m∈R pt care ecuatia (m+3)x²-2(m-1)x+m-2=0,are:
a)2 solutii reale distincte

Answer 1

$\Delta=b^2-4ac= \\ \\ =[-2(m-1)]^2-4(m+3)(m-2)= \\ \\ =4(m-1)^2-4(m^2+m-6)= \\ \\ =4(m^2-2m+1)-4m^2-4m+24= \\ \\ =4m^2-8m+4-4m^2-4m+24= \\ \\ =-12m+28.$

$Pentru~ca~ecuatia~sa~aiba~doua~solutii~reale~si~distincte,~trebuie \\ \\ ca ~\Delta\ \textgreater \ 0. \\ \\ Deci~-12m+28\ \textgreater \ 0. \\ \\ -12m+28\ \textgreater \ 0 \\ \\ -12m\ \textgreater \ -28 \\ \\ 12m\ \textless \ 28 \\ \\ m\ \textless \ \frac{28}{12}= \frac{7}{3}.$

$Deci~m \in \Big(- \infty ; \frac{7}{3} \Big)$