# 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. Is this an error or is there something I’ve missed?

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 $$x=-1$$.