Range of p satisfying quadratic inequality 
What is the set of values of $p$ for which $$p(x^2+2)<2x^2+6x+1,$$ for all real values of $x$?

So I found this one pretty tricky. Feel free to have a look yourself before I outline my thinking.
So I started by taking everything to the RHS forming a quadratic inequality $>0$. Then as long as $2-p$ is positive we know the discriminant must be negative yielding $p<-1$ or $p>\frac{7}{2}$ but we specified $2-p$ is positive so $p<-1$ is the only range of p satisfying the original inequality.
The answer given in the textbook is $-1<p<\frac{7}{2}$. Is this an error or is there something I’ve missed?
 A: Your approach is entirely correct, and seems like the best approach to me as well. I'll write out the details here, if only for myself as I don't have any pen and paper at hand.
Bringing all terms to one side, for each value of $p$ we get a quadratic polynomial in $x$, and we want
$$(2-p)x^2+6x+(1-2p)>0,$$
for all $x$. This happens if and only if $2-p>0$ and the discriminant of the quadratic is negative, i.e.
$$(-6)^2-4(2-p)(1-2p)<0.$$
The latter is a quadratic in $p$, and expanding the products yields
$$-4(2p^2-5p-7)<0.$$
By the quadratic formula we see that this inequality holds if and only if
$$p<\frac{5-\sqrt{25+56}}{4}=-1\qquad\text{ or }\qquad p>\frac{5+\sqrt{25+56}}{4}=\frac72.$$
We already found the necessary condition that $2-p>0$, so together this shows that
$$p(x^2+2)<2x^2+6x+1,$$
if and only if $p<-1$. It seems that you are correct, and your textbook is wrong.

Just as a sanity check; plugging in $p=0$ yields the inequality
$$0<2x^2+6x+1,$$
which should hold for all $x$ according to the solution in your textbook. But this inequality fails for $x=-1$.
