Finding the value of a variable in a quadratic

Question:

The number of negative integral values of $m$ for which the expression $x^2+2(m-1)x+m+5$ is positive $\forall$ $x>1$ is?

For me, solving this question if the parameter "$\forall$ $x>1$" was not given would be quite easy. But how do I solve it under the given parameter?

There are two ways for having $\;p(x)=x^2+2(m-1)x+m+5>0\enspace\forall x>1$:

• either $p(x)$ has no real root, which means the reduced discriminant $\Delta'=(m-1)^2-(m+5)=(m+1)(m-4)<0$ is negative;
• or $p(x)$ has real roots, say $\xi_0\le \xi_1$ , i.e. $\Delta'\ge 0$. In this case, the condition means $1$ is on the right of the interval of the roots. This is satisfied if and only if $p(1)\ge 0$ ($1$ is outside of the interval of the roots or is a root) and, using Vieta's relations, $1>\frac12(\xi_0+\xi_1)=1-m$.

1. $f(1) >0$
2. $\displaystyle -\frac{b}{2a}<0$

Symbols have their usual meaning