If f(x)=px^2+2px-5 , (x is real) what are the requirements for f(x), to be f(x)<0 ?
Here we can simplify the function as, f(x)=p(x^2+2x+5/p) =p[(x+1)^2 - { (p+5)/5 ]
so is the requirement for f(x) to be,f(x)<0 is???
p<0
or (x+1)^2 <[(p+5)/p] ??
we have to solve $$px^2+2px-5<0$$ if $p=0$ then your inequality is true. Now we consider two cases: A) $$p>0$$ then we have $$x^2+2x-\frac{5}{p}<0$$ and this is equivalent to $$(x+1)^2<1+\frac{5}{p}$$ and this can be written as $$|x+1|<\sqrt{1+\frac{5}{p}}$$ for $$x\geq -1$$ we have $$x<-1+\sqrt{1+\frac{5}{p}}$$ in the other case we get $$x>-1-\sqrt{1+\frac{5}{p}}$$ for B) $$p<0$$ then we have $$(x+1)^2>1+\frac{5}{p}$$ if $$1+\frac{5}{p}\le 0$$ then our inequality is true. in the othere case we get $$|x+1|>\sqrt{1+\frac{5}{p}}$$ can you proceed?
By completing the square, $$px^2+2px-5=p(x+1)^2-5-p.$$
Then $p$ may certainly not be positive (as $(x+1)^2$ is unbounded) and the expression achieves the global maximum $-5-p$ for $x=-1$. This is negative for
$$-5