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Find the arithmetic mean

  1. Nov 1, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If four distinct points on the curve [itex]y=2x^4+7x^3+3x-5[/itex] are collinear, then find the arithmetic mean of x-coordinates of the aforesaid points.

    2. Relevant equations

    3. The attempt at a solution
    I think that the four points mentioned must be the roots of the equation.
     
  2. jcsd
  3. Nov 1, 2012 #2

    SammyS

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    Graph the function. It only has one real root.
     
  4. Nov 1, 2012 #3
    Are you sure? I thought for any polymonial, complex roots come in even numbers, so that quartics have 0, 2 or 4 real roots (some may be repeated, mind)
     
  5. Nov 1, 2012 #4

    Curious3141

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    There are 2 real roots.
     
  6. Nov 1, 2012 #5

    Ray Vickson

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    I assume you mean that for f(x) = 2x^4+7x^3+3x-5, the four points (x1,f(x1)), (x2,f(x2)),(x3,f(x3)), (x4,f(x4)) lie on a straight line. If you plot f(x) you will see that the places where you can have 4 collinear points on the curve y = f(x) are quite limited (although there are still infinitely many possibilities). I don't see why the roots of f(x) have much to say about this problem.

    RGV
     
    Last edited: Nov 1, 2012
  7. Nov 1, 2012 #6

    SammyS

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    I assume that since this thread is posted in the Precalculus Mathematics section, you are not to use calculus.

    Is that correct?
     
  8. Nov 1, 2012 #7

    utkarshakash

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    A liitle bit of it won't do any harm.
     
  9. Nov 1, 2012 #8

    utkarshakash

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    I plotted the graph and it comes out that there are only 2 real roots. So the points mustn't be the roots. Then what are those 4 points?
     
  10. Nov 1, 2012 #9

    SammyS

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    Inflection points for [itex]y=2x^4+7x^3+3x-5[/itex] occur where the second derivative is zero.

    These occur at (-7/4, -3713/128) and at (0, -5).

    The line passing through these points has the equation, y = (439/32)x-5 .

    Subtracting (439/32)x-5 from your function gives, [itex]\displaystyle y=2 x^4+7 x^3-\frac{343}{32}x\ .[/itex]

    Here's the graph of that from WolframAlpha.

    attachment.php?attachmentid=52544&stc=1&d=1351829819.gif

    Of course two of the zeros are at the inflection points. You can then find the other two for the remaining quadratic.
     

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  11. Nov 1, 2012 #10

    ehild

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    The four points are on a straight line y=ax+b, and are points of the curve at the same time. That means: [itex]ax+b=2x^4+7x^3+3x-5[/itex]
    The x coordinates of the collinear points are roots of the rearranged equation [itex]2x^4+7x^3+(3-a)x-5-b=0[/itex], and the equation has four real roots.

    You need the geometric mean of the x coordinates, which is the sum of roots divided by 4. And you know the relation between roots and coefficients.

    ehild
     
  12. Nov 2, 2012 #11

    utkarshakash

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    You are a genius. Thanks for helping. There was no need of those calculus methods which were extremely complicated.
     
  13. Nov 2, 2012 #12

    utkarshakash

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    Can you please help me on the other question of mine posted by the title
    'Find least value of a'?
     
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