# Finding set of values using inequalities

I'm attempting a question in my math book (self teaching so don't have a personal tutor to ask). I'm getting confused as to what I'm supposed to be doing. Here's the question: What is the set of values of p for which $p(x^2 + 2) < 2x^2 + 6x + 1$ for all real values of x?

I've rearranged this to the following inequality: $(p - 2)x^2 - 6x + 2p - 1 < 0$

Now I've reworded the question to "find values of p for which $(p - 2)x^2 - 6x + 2p - 1 < 0$ for all real values of x. Because the inequality is $< 0$, I'm guessing that I'm seeking all the parabolas that are below the x-axis. I'm already getting a bit lost as you might tell. Quadratic equations that don't cross the x-axis do not have real roots, so I'm looking at the discriminant $'b^2 - 4ac' < 0$ where $a = (p - 2)$, $b = -6$, and $c = (2p - 1)$. This boils down to the inequality $2p^2 - 5p - 7 > 0$, the solutions of which I compute to be $p < -1$ or $p > \frac{7}{2}$.

The answer in the back of the book is $-1 < p < \frac{7}{2}$ which is not the same as what I have but has the same boundary conditions.

My request is for someone to help me understand each step of the solution please. I'm getting a bit lost and losing the intuition of it all.

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The book answer is wrong. If you set p=0, x=-1.5 the inequality is not satisfied. As Robert Israel says, you cannot have p>2 or for large positive $x$ the inequality will fail. –  Ross Millikan Apr 21 '11 at 19:54
If you quoted the question correctly, the book is wrong. This quadratic is always positive if $p > 7/2$, and always negative if $p < -1$.