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**harrychess****Member**- Registered: 2014-04-04
- Posts: 33

Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18.

How do you work this problem?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Hi bobbym;

What does

```
p \[Element]
Reals && ((x < -4 && p <= (-9 + 4 x + x^2)/(8 + 2 x)) ||
x == -4 || (x > -4 && p >= (-9 + 4 x + x^2)/(8 + 2 x)))
```

mean?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Check post #2. That code is incomplete.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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How did you do that?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

The experimental way of course.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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I hope you do not mean plugging in random numbers!

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Hahhahahahhaahhahahah! Oh Mr. Chattopadhyay you are in rare form today. Thank you, I needed to laugh and that was funny.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

I am getting p=-5 with the p=1 bobbym got,

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Very good answer!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

Are you getting any other possibility?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

I am looking for something else but so far no.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

Agnishom wrote:

I hope you do not mean plugging in random numbers!

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```
In[1]:= 2 (x + 4) (x - 2 p)
Out[1]= 2 (4 + x) (-2 p + x)
In[2]:= \!\(
\*SubscriptBox[\(\[PartialD]\), \(x\)]\((2\ \((4 +
x)\)\ \((\(-2\)\ p + x)\))\)\)
Out[2]= 2 (4 + x) + 2 (-2 p + x)
In[3]:= Solve[% == 0, x]
Out[3]= {{x -> -2 + p}}
In[4]:= 2 (4 + x) (-2 p + x) /. {{x -> -2 + p}}
Out[4]= {2 (-2 - p) (2 + p)}
In[8]:= Solve[2 (-2 - p) (2 + p) == -18, p]
Out[8]= {{p -> -5}, {p -> 1}}
```

For harrychess:

Let f(x) = 2 (x + 4) (x - 2 p)

Let f'(x) be the derivative of f(x) with respect to x [,i.e, 2 (4 + x) + 2 (-2 p + x)]

Let g(p) be a polynomial in terms of p which satisfies the equation f'(x) = 0 [,i.e, x = p - 2]

Find p such that 2 (g(p) + 4) (g(p) - 2p) = -18 [,i.e, Solve this, 2 ( 4+ (p-2)) ((p-2)-2p)=-18]

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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Hi bobbym,

Is the above experimental enough?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

I did it in a similar way but there is a shorter way to do the problem I think.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,510

I found solutions for these:

p=-7

p=-5

p=1

p=3

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Hi phrontister;

Did you test that in geogebra?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,510

No, I didn't think of it.

So now I've entered the x and p coordinates into G and got the result as per image, but I don't know what to do next.

A straight-line connector isn't going to give any solutions, which I suppose means that I need to fit a curve, probably a separate one for ABC and DEF (I'm only guessing). However, I tried that and failed, except with 3-point circles through ABC and DEF...from which I don't get any other solutions either.

In fact, I think there are no non-integer solutions, and my spreadsheet seems to indicate that my integer solutions are the only answers to this puzzle.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Hi phrontister;

There are only 2 answers for p.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,510

Hi Bobby,

I got 4 answers for p (ie, -7, -5, 1 and 3):

p = -7 and x = -13.....2(-13+4)(-13-2*-7) = -18

p = -7 and x = -5.......2(-5+4)(-5-2*-7) = -18

p = -5 and x = -7.......2(-7+4)(-7-2*-5) = -18

p = 1 and x = -1........2(-1+4)(-1-2*1) = -18

p = 3 and x = -3........2(-3+4)(-3-2*3) = -18

p = 3 and x = 5.........2(5+4)(5-2*3) = -18

Did I misunderstand something about the question, or make an error?

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Hi;

You can check these quickly with geogebra.

When p = - 7 the minimum is - 50

When p = 3 the minimum is - 50

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,510

Oops! I overlooked the word 'minimum! >facepalm<

Yes...there are only solutions 1 and -5.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,751

Misheeru is the facepalm kid. I can not deal with two of ya. My brain is exploding.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,510

Yes, I learnt that term from Misheeru, and I felt the need to use it now!

I've been trying to use G with this, but am getting nowhere. So far I've entered the two {x,p} coordinates and drawn a line through them, but I've run out of ideas....and I may be on the wrong track doing that, anyway.

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