Question

Buna! Imi explicati cum se face ex 2, va rog?

Answer 1

[tex]x^2-7x+10 \geq 0 =\ \textgreater \ \left(x-5\right)\left(x-2\right)\ge \:0 \\ \\ \\(x-5) = 0 =\ \textgreater \ x = 5 ; (x-2) = 0 =\ \textgreater \ x=2 \\ \\ \\ \boxed{x= 5} \\ \boxed{x= 2} \\x\le \:2\quad \mathrm{sau}\quad \:x\ge \:5 \\Am atasat tabelul[/tex]

Answer 2

x²-7x+10≥0
inecuatie de gradul II
a =1;  b =-7  si c =10
Δ=b²-4ac=49-4×10=49-40=9
Δ = 9
x₁=(-b-√Δ)/2a=2
x₂(-b-√Δ)/2a=5

a > 0(a=1)  si Δ> 0 (Δ = 9)
deci inecuatia are solutiile:
x∈(-∞; x₁] ∪[ x₂; +∞)
x∈(-∞; 2] ∪[ 5; +∞)