# Need help with inequality problem

1. Jun 4, 2015

### Theodore Hodson

1. The problem statement, all variables and given/known data
What is the set of values of p for which p(x^2+2) < 2x^2+6x+1 for all real values of x?

2. Relevant equations

p(x^2+2) < 2x^2+6x+1

3. The attempt at a solution
I know I need to use my knowledge of the discriminant here, but the fact that its an inequality is confusing me somewhat. Would be very grateful if anyone could point me in the right direction :)

TYVM ! :)

Last edited by a moderator: Jun 4, 2015
2. Jun 4, 2015

### BiGyElLoWhAt

Can you simply solve for p using long division?

3. Jun 4, 2015

### SammyS

Staff Emeritus
Yes. The discriminant can be a big help.

In general, what is the discriminant used for?

Last edited by a moderator: Jun 4, 2015
4. Jun 4, 2015

### Theodore Hodson

umm its used to give us information about the roots of a quadratic. Like whether the roots are real or not .Just not quite sure how I can apply it here. I can see that because the question is asking about 'real' values of x its probably wanting me to use this 'b^2-4ac>=0 but I'm still pretty stumped.

5. Jun 4, 2015

### SammyS

Staff Emeritus
Yes.

Roots of a quadratic : values of x that make the quadratic equal to zero, Right?

Therefore, subtract p(x2 +2) from both sides of the inequality.

Now, if it was an equation rather than an inequality, what are the various outcomes that the discriminant predicts?

6. Jun 4, 2015

### Theodore Hodson

Right I think I figured it out. But I still got a problem -I'll talk about that in a sec. So going from what you said I subtracted p(x^2+2) from both sides and ended up with:

0< (2-p)x^2 +6x +1 -2p

Okay I've got a quadratic inequality here now. So because the question is asking for p values for all 'real' values of x then the roots of the quadratic for those 'real' values of x will be 'real' too. Which means we can use the discriminant to create an inequality to show when the quadratic has real roots and then isolate p in that inequality.

So using the discriminant (2-p)x^2+6x+1-2p has real roots when

(6)^2-(4)(2-p)(1-2p)>= 0

which simplifies to:

-8p^2+20p+28>=0

Factoring this result, we get:

-4(2p-7)(p+1)>=0

So solving this inequality, p is greater than or equal to zero within this range -1<=p<=7/2.

Okay so I think kinda understand it tho I'm still a bit confused. Checked my answer in the textbook and its wrong. It's meant to be -1<p<7/2. Can you show where I went wrong?? much appreciated :)

7. Jun 4, 2015

### Ray Vickson

No: you have it exactly backwards: the graph of the function $y = (2-p) x^2 + 6x + 1 - 2p$ must lie in the region $y > 0$ for all $x$. That means that the equation $(2-p) x^2 + 6x + 1 - 2p = 0$ must have NO real roots. What is the condition for that?

8. Jun 4, 2015

### Theodore Hodson

Aha! I see now. Man I'm such a dumbass. Perfect example of me not using common sense. So if you were to graph that it would never intersect the x-axis. Alright so the last step in my working would change to
-4(2p-7)(p+1)<0 which would mean p is less than zero within this range -1>p>7/2. Thanks a lot!!!

I think the thing that really got me confused was it asking for values of p that satisfy the inequality for all 'real' values of x. What does 'for all real values of x' mean in the context of the question?

9. Jun 4, 2015

### Ray Vickson

It means that the graph of $y = p(x)$ lies either in the region $y > 0$ or the region $y < 0$. That means that the function $p(x)$ cannot cross or touch 0, and that, in turn, means that the equation $p(x) = 0$ has no real roots.

Think about what it would mean if the equation $p(x) = 0$ DID have one or more real roots. Could the inequality $p(x) > 0$ or $p(x) < 0$ then hold for ALL $x$?

10. Jun 5, 2015

### Theodore Hodson

Ahh right. If they did have real roots then the inequalities p(x)>0 and p(x)<0 wouldn't hold for all x cause the functions would intersect the x-axis at some x value. Thank-you for the help sir :)

11. Jun 5, 2015

### SammyS

Staff Emeritus
That inequality, -1 > p > 7/2, is not a valid inequality. It says p is less than -1 and p is greater than 7/2. I doubt that you actually mean that.

Once you get that sorted out ...

You are not quite done solving this problem.

What you have found so far is those values of p such that $\ y=(2-p)x^2+6x+1-2p \$ doesn't cross the x-axis, so $\ (2-p)x^2+6x+1-2p \$ is always positive or always negative. You only want one of those.

Which one will give you $\ p\,(x^2+2) < 2x^2+6x+1 \$ for all $\ x\ ?$

12. Jun 5, 2015

### Theodore Hodson

Yeah you're totally right I didn't mean -1>p>7/2 I meant -1<p<7/2 which is in fact the answer in the back of the book. Though having gone through the question again I think -1<p<7/2 isn't right either so the book must have made an error. So let me work things through from the start. Please tell if my thinking is wrong I really appreciate it :)

Alright so going back from 2-px^2+6x+1-2p>0 - this would be a parabola that is totally above the x-axis (so always positive) and if you move everything to the other side (p-2x^2-6x+2p-1<0) then this would be a parabola totally below the x-axis (so always negative). So both have no real roots. I'll use the one that is always positive and therefore has no real roots. So if it has no real roots then (using the discriminant of this always positive quadratic) (6)^2-4(2-p)(1-2p) must be less than 0. The brackets expand to make:

-8p^2+20p+28<0
This quadratic would then be visualised as a 'sad' face parabola and now that we've got a quadratic inequality in p we can isolate p and find the range of values it can take.

Then factoring we get:

-4(2p-7)(p+1)<0
dividing by -4

(2p-7)(p+1)<0

So this quadratic would have roots at -1 and 7/2 and since its an 'upside-down' parabola it means that it is less than zero when p>7/2 and p<-1. So from this we've figured out the range of values that P can take.

Have I got this right now?

13. Jun 5, 2015

### haruspex

Yes. You can easily check a few values, like p=0.

14. Jun 5, 2015

### SammyS

Staff Emeritus
Sort of right.

It's p < -1 OR p > 7/2 so that the discriminant is negative, i.e. there are no zeros.

But as I pointed out in a previous post:

In other words:

For the any particular value of p such that $\displaystyle\ p\in(-\infty,\,-1)\cup(7/2,\,\infty)\,,\$ one of the following is true.

Either $\ (2-p)x^2+6x+1-2p>0 \$ for all $\ x\$ or $\ (2-p)x^2+6x+1-2p<0 \$ for all $\ x\$.

You need to pick only those values of p that give $\ (2-p)x^2+6x+1-2p>0 \$. Right?

15. Jun 6, 2015

### Theodore Hodson

11

16. Jun 6, 2015

### Theodore Hodson

Ahh right okay I think I understand. I think I was getting massively confused before. Let me go from near the beginning again. Thanks for bearing with me all this time :)

So we have $y=(2-p)x^2+6x+1-2p$ and using the discriminant, I've found those values of $p$ for which

$y=(2-p)x^2+6x+1-2p$ has no real roots.

i.e those values of $p$ for which $(2-p)x^2+6x+1-2p>0$ OR $(2-p)x^2+6x+1-2p<0$.

And I want to find the range of values of $p$ which will make $(2-p)x^2+6x+1-2p>0$.

So $(2-p)x^2+6x+1-2p$ will be greater than zero or less than zero depending on whether $p>\frac{7}{2}$ or $p<-1$.

Okay so lets find the y-intercept of $(2-p)x^2+6x+1-2p$ for a $p>\frac{7}{2}$ and the y intercept of $(2-p)x^2+6x+1-2p$ for a $p<-1$.

Depending upon whether the y-intercept is negative or positive it will show us if this range of $p$-values gives us the totally positive or totally negative version of $(2-p)x^2+6x+1-2p$.

Alright so y intercept of $(2-p)x^2+6x+1-2p$ when $p>\frac{7}{2}$. Okay so let's choose a $p$ value of 3.6 which is $>\frac{7}{2}$.
Doing the calculations this gives us a y-intercept of -6.2.

Okay y intercept of $(2-p)x^2+6x+1-2p$ when $p<-1$. Choosing a $p$ value of -2 which is $<-1$. Doing the calculations gives us a y-intercept of 5.

So concluding from the above, $(2-p)x^2+6x+1-2p>0$ if $p<-1$. And $(2-p)x^2+6x+1-2p<0$ if $p>\frac{7}{2}$.

We want to find the set of values of $p$ for which $(2-p)x^2+6x+1-2p>0$ which is just if $p<-1$.

17. Jun 6, 2015

### SammyS

Staff Emeritus
Yes, your final answer is $\ p<-1\$ .

Comment:
What you've done by using test points to check on the y-intercept is fine. However, ...

You could be a little more thorough in making sure that ALL of the values for $\ p<-1\$ make the y-intercept positive and also making sure that ALL of the values for $\ p>7/2\$ make the y-intercept negative. (Actually, these facts should be readily apparent simply by inspection.)

If $\ p<-1\,,\$ then multiplying by -2 we know $\ -2p>2\$ . Adding 1 gives $\ 1-2p>3\$ . So that's fine.

You can handle $\ p>7/2\$ similarly.

Have also made an avi video of a simulation in which p is increased from -3 to 4.75 in steps of 0.25 .

I'm not sure how to upload that.