Quadratic Inequalities

harrychess · 2014-04-04 13:19:48

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

How do you work this problem?

bobbym · 2014-04-04 13:47:26

Hi;

Agnishom · 2014-04-04 13:53:46

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?

bobbym · 2014-04-04 13:55:27

Check post #2. That code is incomplete.

Agnishom · 2014-04-04 13:57:27

How did you do that?

bobbym · 2014-04-04 13:57:56

The experimental way of course.

Agnishom · 2014-04-04 14:02:36

I hope you do not mean plugging in random numbers!

bobbym · 2014-04-04 14:04:03

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

ShivamS · 2014-04-04 14:12:41

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

bobbym · 2014-04-04 14:13:06

Very good answer!

ShivamS · 2014-04-04 14:14:24

Are you getting any other possibility?

bobbym · 2014-04-04 14:17:35

I am looking for something else but so far no.

ShivamS · 2014-04-04 14:18:26

Agnishom wrote:

I hope you do not mean plugging in random numbers!

Agnishom · 2014-04-04 19:36:50

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]

Agnishom · 2014-04-05 12:57:54

Hi bobbym,

Is the above experimental enough?

bobbym · 2014-04-05 13:33:38

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

phrontister · 2014-04-05 15:13:25

I found solutions for these:

p=-7
p=-5
p=1
p=3

bobbym · 2014-04-05 15:49:06

Hi phrontister;

Did you test that in geogebra?

phrontister · 2014-04-06 02:11:30

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.

bobbym · 2014-04-06 03:17:19

Hi phrontister;

There are only 2 answers for p.

phrontister · 2014-04-06 04:07:43

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?

bobbym · 2014-04-06 04:12:49

Hi;

You can check these quickly with geogebra.

When p = - 7 the minimum is - 50

When p = 3 the minimum is - 50

phrontister · 2014-04-06 04:25:08

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

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

bobbym · 2014-04-06 04:29:19

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

phrontister · 2014-04-06 04:38:27

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.

Math Is Fun Forum

#1 2014-04-04 13:19:48

Quadratic Inequalities

#2 2014-04-04 13:47:26

Re: Quadratic Inequalities

#3 2014-04-04 13:53:46

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#4 2014-04-04 13:55:27

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#5 2014-04-04 13:57:27

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#6 2014-04-04 13:57:56

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#7 2014-04-04 14:02:36

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#8 2014-04-04 14:04:03

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#9 2014-04-04 14:12:41

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#10 2014-04-04 14:13:06

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#11 2014-04-04 14:14:24

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#12 2014-04-04 14:17:35

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#13 2014-04-04 14:18:26

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#14 2014-04-04 19:36:50

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#15 2014-04-05 12:57:54

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#16 2014-04-05 13:33:38

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#17 2014-04-05 15:13:25

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#18 2014-04-05 15:49:06

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#19 2014-04-06 02:11:30

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#20 2014-04-06 03:17:19

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