Răspuns:
Ca f(x) sa aibe valori pozitive => a>0 si delta<=0
[tex]f(x)=ax^2+6(a+1)x+9a+31[/tex]
[tex]a=a;b=6(a+1);c=9a+31[/tex]
Prima conditie: a>0
A doua:
[tex]Delta\leq 0[/tex]
[tex](6a+6)^2-4a(9a+31)\leq 0[/tex]
[tex]36a^2+72a+36-36a^2-124a\leq 0[/tex]
[tex]-52a+36\leq 0\\[/tex]
[tex]-52a\leq -36[/tex]
[tex]52a\geq 36=>a\geq \frac{9}{13}[/tex]
[tex]=> a[/tex]∈[tex][\frac{9}{13};+inf)[/tex]
