How to find the range of p where p(x^2 + 2) < 2x^2 + 6x + 1

[Masaru]

I have a problem with the following question:

Q: What is the set of values of p for which p(x^2+2) < 2x^2+6x+1 for all real values of x?

My working is as follows:

First, just expand the left-hand side of the inequality,
px^2 + 2 p <

2x^2 + 6x + 1
Then, rearrange it to:
(2 - p)x^2 + 6x + 1 - 2p > 0

The determinant b^2 - 4ac = 6^2 - 4(2 - p)(1 - 2p)
= 36 - 4(2p^2 - 5p + 2)
= - 8 p^2 + 20 p + 28 >(or =) 0
So
2p^2 - 5p - 7 < (or =) 0
So
(2p - 7)(p + 1) < (or = )0
Therefore, - 1 < (or=) p < (or=) 7/2

However, I know that this working cannot be right because this is just a working to determine the range of p where x are real values, and it does not really mean that the left-hand side of the inequality is always less than the right-hand side of it.

In fact, when p = 1, the left-hand side would be x^2 + 2, and you could clearly see that x^2 + 2 is not always below the curve of

2x^2 + 6x + 1 ​

if you check it using a graphing calculator.

And the answer provided by the text says that the right answer is:

p < - 1
&
p > 7/2

It looks like this answer comes from exactly the same way as I did as above (the differences are that it does not have "=" sign, which I do not understand why, and the range is totally opposite (complement) of my answer).

Having said that, this answer provided in the textbook does not seem to be correct, either.

Because, for example, when p = 4, which is more than 7/2, the left-hand side of the inequality would be 4x^2 + 8, and you could clearly see that it is actually more than

2x^2 + 6x + 1 ​

all the time for the same value of x, not less than the right-hand side of the inequality!

I have been very confused about this question as well as the answer provided in the book, and I would much appreciate it if someone can help me to find out the right solution for this question.


(By the way, we are not allowed to use a graphing calculator for this question. I just used the online graphing calculator to check both my and the book's answers)
​

 

[Dr.Peterson]

When you used the discriminant, you required it to be non-negative, which implies that the original inequality will be an equality for some real values of x. That is exactly the opposite of what you need! You want it NEVER to be an equality, so that it will ALWAYS be an inequality (for all x). That is why your answer is the complement of the correct answer.

But you are right to check. The discriminant alone does not tell you whether the inequality that is always true is the one they are asking about, or the opposite! Possibly they forgot to check this themselves.

Taking your rearranged inequality, (2 - p)x^2 + 6x + 1 - 2p > 0, as you note, when p = 4, we have

-2x^2 + 6x - 7 > 0

which is NEVER true; the left side is a parabola that opens downward, with its vertex below the axis. On the other hand, when p = -2, the inequality is

4x^2 + 6x + 5 > 0

which is always true. So the correct solution is p < -1 only.

The score is 1 for the author, and 1 for you. Good work, actually!

 

[mmm4444bot]

What is the set of values of p for which p(x^2+2) < 2x^2+6x+1 for all real values of x?


… the answer provided by the text says that the right answer is:


p < - 1
&
p > 7/2

You need to consider cases separately (p>0 and p<0), and that book answer is not correct.

We know that the parabola of 2x^2 + 6x + 1 opens upward. Checking the vertex, at x=-b/(2a), we find a negative y-coordinate. Therefore, this parabola dips below the x-axis.

Now, when p is positive, the parabola of px^2 + 2p also opens upwards.

Therefore (for positive p), if px^2 + 2p is to be less than 2x^2 + 6x + 1 for all x, then its parabola must lie entirely below the other. (And remember, the other has a vertex below the x-axis). Well, consider the shape and position of the px^2+2p parabola, as positive p values approach zero. As positive p gets smaller and smaller, the parabola is shifting down and flattening out. But it will never lie below the x-axis, for any positive p, because 2p is its y-intercept, yes? (Indeed, when p reaches zero, the graph is no longer a parabola; it has become so flat that it's now a line -- the x-axis itself). So the upward-opening parabolas from positive p values will never lie entirely below the parabola of 2x^2+6x+1 (which dips below the x-axis). This demonstrates that there are no positive p solutions to the original inequality.

Also, when you used the Discriminant from the quadratic polynomial (2 - p)x^2+6x+(1 - 2p), that just tells you p values for which this parabola has Real x-intercepts. This quadratic gives the distance between the two parabolas discussed above. I don't think there's a correspondence between the original inequality's solutions for p and the values of p for which the distance parabola has Real roots. It's actually giving the p-values for which the two parabolas touch at only one point. That fact is helpful.

Once we realize, from the analysis above, that p must be negative, then it's just a matter of determining whether p is greater than or less than -1 (the value for which the two parabolas touch at only one point). For that, you would consider cases. You don't want px^2+2p to be equal to 2x^2+6x+1, so p must be negative but not -1. If you're able to find a solution for px^2+2p=2x^2+6x+1, when testing cases for p, then that means the upward- and downward-opening parabolas intersect, so one cannot be less than the other.

If my post seems confusing, then I would suggest that you investigate the shifting positions of the p*(x^2+2) parabola, by graphing several values of p (like, -3, -2, -1, -1/2, 0, 1, 2, 3, 4, 5), along with the graph of 2x^2+6x+1. Study what's happening, and then try re-reading my post. :cool:

 

[mmm4444bot]

I had to edit a mistake in my post, so I decided to add a graph showing 2x^2+6x+1 in red and p(x^2+2) in blue. I let p take on integer values from -8 to 8.

I hope the picture helps you to visualize the analyses I discussed earlier. You posted a good exercise, for practicing what we've learned about how the parameters affect graphs of quadratic polynomials (eg: leading coefficients and y-intercepts) and shifting functions in general.

 

[Masaru]

Now I see where I went wrong!

When you used the discriminant, you required it to be non-negative, which implies that the original inequality will be an equality for some real values of x. That is exactly the opposite of what you need! You want it NEVER to be an equality, so that it will ALWAYS be an inequality (for all x). That is why your answer is the complement of the correct answer.

Now I see where I went wrong.

A part of my working was:

- 8 p^2 + 20 p + 28 >(or =) 0
​

But the sign should be "<".

The reason that I used > (or =) is that I simply misread the question and I got muddled up about what the question says - "for real value of x". I was just thinking that x has to be a real number even though, as you say, this is not equality, but inequality.

So I worked this question again from the scratch as follows:

For px^2 + 2p < 2x^2 + 6x + 1 to be true,

1) There should be no point of intersection between both curves represented by each side of this inequality.

2)

px^2 + 2p​

should be always below the curve of

2x^2 + 6x + 1 ​

for any value of x

For 1), rearrange and change the inequality to (2 - p)x^2 + 6x + 1 - 2p = 0

Discriminant b^2 - 4ac has to be less than 0 here because there should not be any solution for x to satisfy the equation so that there is no point of intersection.

Therefore,

- 8 p^2 + 20 p + 28 < 0
So
2p^2 - 5p - 7 > 0
So
(2p - 7)(p + 1) > 0
Therefore, p > 2/7 or p < - 1

But when p >2/7,

px^2 + 2p ​

​

>​

​

2x^2 + 6x + 1
(e.g. when p = 5, x = 0, the leff-hand side > right-hand side)

So, for

px^2 + 2p ​

​

​

<​

​

2x^2 + 6x + 1 to be true,

​

p < - 1 only​

​

Thank you again for your assistance

 

[Masaru]

Thank you for your detailed analysis.

You need to consider cases separately (p>0 and p<0), and that book answer is not correct.

We know that the parabola of 2x^2 + 6x + 1 opens upward. Checking the vertex, at x=-b/(2a), we find a negative y-coordinate. Therefore, this parabola dips below the x-axis.

Now, when p is positive, the parabola of px^2 + 2p also opens upwards.

Therefore (for positive p), if px^2 + 2p is to be less than 2x^2 + 6x + 1 for all x, then its parabola must lie entirely below the other. (And remember, the other has a vertex below the x-axis). Well, consider the shape and position of the px^2+2p parabola, as positive p values approach zero. As positive p gets smaller and smaller, the parabola is shifting down and flattening out. But it will never lie below the x-axis, for any positive p, because 2p is its y-intercept, yes? (Indeed, when p reaches zero, the graph is no longer a parabola; it has become so flat that it's now a line -- the x-axis itself). So the upward-opening parabolas from positive p values will never lie entirely below the parabola of 2x^2+6x+1 (which dips below the x-axis). This demonstrates that there are no positive p solutions to the original inequality.

Yes, I can perfectly follow what you are saying here even without drawing graphs.
Thank you very much for your detailed explanation.
Also now I see that one of the solutions in the book is wrong as you say!

 

[Dr.Peterson]

Now I see where I went wrong.

A part of my working was:

- 8 p^2 + 20 p + 28 >(or =) 0
​

But the sign should be "<".

The reason that I used > (or =) is that I simply misread the question and I got muddled up about what the question says - "for real value of x". I was just thinking that x has to be a real number even though, as you say, this is not equality, but inequality.

px^2 + 2p

The important word, which you have omitted here, is "for all real values of x". If the inequality is to be true for all real x, then the equation must be true for no real values of x! That seems to be what you missed, more specifically.

So I worked this question again from the scratch as follows:

For px^2 + 2p < 2x^2 + 6x + 1 to be true,

1) There should be no point of intersection between both curves represented by each side of this inequality.

2) should be always below the curve of 2x^2 + 6x + 1
for any value of x

For 1), rearrange and change the inequality to (2 - p)x^2 + 6x + 1 - 2p = 0

I think it is important to keep the inequality: (2 - p)x^2 + 6x + 1 - 2p < 0. If you have solved quadratic inequalities before, you know that the solution (for fixed p) will be all real numbers if there are no places where the left side is 0, and you can show that it is negative for any one value of x, since it can't change sign anywhere. That is the basis of what you say next:

Discriminant b^2 - 4ac has to be less than 0 here because there should not be any solution for x to satisfy the equation so that there is no point of intersection.

Therefore,

- 8 p^2 + 20 p + 28 < 0
So
2p^2 - 5p - 7 > 0
So
(2p - 7)(p + 1) > 0

Therefore, p > 2/7 or p < - 1

You meant 7/2, of course.

But when p >2/7, px^2 + 2p > 2x^2 + 6x + 1

(e.g. when p = 5, x = 0, the leff-hand side > right-hand side)

So, for

px^2 + 2p < 2x^2 + 6x + 1 to be true,

p < - 1 only

I'm not quite happy with this, because you have (following my lead) only shown what happens for a particular value of p. How do we know that is sufficient to prove that this is true for all p < -1, and not for any p > 7/2?

Any thoughts? Some of my comments above may be relevant; or you may consider the sign of the leading coefficient, and how it affects the graph of a parabola.

 

[Masaru]

How to prove that answer is p < - 1 only

You meant 7/2, of course.

I'm not quite happy with this, because you have (following my lead) only shown what happens for a particular value of p. How do we know that is sufficient to prove that this is true for all p < -1, and not for any p > 7/2?

Any thoughts? Some of my comments above may be relevant; or you may consider the sign of the leading coefficient, and how it affects the graph of a parabola.​

Sorry, but I meant "7/2", not "2/7".

What I have understood so far is that the range of p has to be either

p > 7/2​

or

p < -1​

so that there is no point of intersection between the two parabolas. However, this is only one of the conditions that you need to satisfy, and as you explained to me previously, it does not necessarily mean that

px^2 + 2p ​

< ​

2x^2 + 6x + 1 ​

and the sign can be the other way around (i.e. >).

So I would use the previous analysis from mmm4444bot to say that p cannot be positive (p < 0) because you would always get a point of intersection or

px^2 + 2p ​

>​

2x^2 + 6x + 1 ( the sign of inequality would become the opposite to what we want) if p were positive.​

Therefore, we need to exclude p > 7/2 and that p < - 1 should be the only answer.

I made a bit of short-cut in my explanation above and did not include the explanation from mmm4444bot,
but isn't it enough to prove that the answer is p >7/2 only, Dr Peterson?​

 

[Dr.Peterson]

So I would use the previous analysis from mmm4444bot to say that p cannot be positive (p < 0) because you would always get a point of intersection or

px^2 + 2p ​

>​

2x^2 + 6x + 1 ( the sign of inequality would become the opposite to what we want) if p were positive.​

Therefore, we need to exclude p > 7/2 and that p < - 1 should be the only answer.

I made a bit of short-cut in my explanation above and did not include the explanation from mmm4444bot,
but isn't it enough to prove that the answer is p >7/2 only, Dr Peterson?

Yes, you can do that; I was hoping you would add a concise version of something of the sort to your answer, to make it clear that you fully understand it.

I would do something like this: In the simplified form, (2 - p)x^2 + 6x + 1 - 2p > 0, if 2-p were negative, the graph could not lie entirely below the x-axis, because of its end behavior (becoming positively infinite for large x), so in order to be true for all x, it is necessary to have 2 - p > 0, so p < 2. This is true for the case p < -1, but not for p > 7/2. I think this is a little more precise than the claim that p < 0.