Question

1.F(x)=x^2-2mx+m^2+16 sa fie situat deasupra axei oxei ox, cum pot rezolva prin delta, pls!
2.sa se rezolve inecuatia 3/x^2-2x > sau egal 1

Answer 1

$1) f(x) = x^2-2mx+m^2+16 \\ \\ \text{Ca Graficul functiei sa fie situat deasupra axei Ox, trebuie ca varful sa} \\ \text{fie situat deasupra axei Ox, adica, la} V( \frac{-b}{2a}, \frac{-\Delta}{4a}), \quad \frac{-\Delta}{4a}\ \textgreater \ 0 \\ \\ \Delta = 4m^2-4(m^2+16) =4(m^2-m^2-16) = 4\cdot(-16) = -64 \\ \\ \Rightarrow \frac{-(-64)}{4\cdot 1} = \frac{64}{4} = 16 \ \textgreater \ 0 (A),\forall m\in \mathbb_{R} \Rightarrow m\in \mathbb_{R}$