Question

Sa se determine parametrul real m,astfel incit ecuatia mx^2-2(m+1)x+m-5=0 sa aiba solutii reale.

Answer 1

$mx^2-2(m+1)x+m-5=0 \\\\ a=m\\ b=-2(m+1) \\ c=m-5 \\\\ \hbox{Pentru ca ecuatia sa aiba valori reale punem conditia ca} \ \Delta \ \textgreater \ 0 \\\\\ \Delta=[-2(m+1)]^2-4*m*(m-5) \\\\ \Delta=4(m+1)^2-4m^2+20m \\\\ \Delta= 4m^2+8m+4-4m^2+20m \\\\ \underline{\Delta=28m+4} \\\\ \Delta\ \textgreater \ 0 \\\\\\ 28m+4 \ \textgreater \ 0 \ \ |:4 \\\\ 7m+1\ \textgreater \ 0 \\\\ 7m\ \textgreater \ -1 \\\\ m\ \textgreater \ -\frac{1}{7} \\\\\\ \boxed{m \in (-\frac{1}{7};+ \infty)}$