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Math Help - Finding three X values for y=ax^3+bx^2+cx+d

  1. #1
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    Finding three X values for y=ax^3+bx^2+cx+d

    Hello,
    I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
    It cuts the y-axis at (0,30). The gradient at this point is -1.
    The gradient where x=1, is -17 and the f ``(x) at this point is -10.

    I s.t the coordinates given; set f `(x) =-1 and
    f ``(x)=-10.

    I think d=30 and c=-1.
    So I need to solve for a,b.

    I didn't think solving for x would be this tricky? What have I missed?

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  2. #2
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    y=0 to solve

    Quote Originally Posted by Neverquit View Post
    Hello,
    I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
    It cuts the y-axis at (0,30). The gradient at this point is -1.
    The gradient where x=1, is -17 and the f ``(x) at this point is -10.

    I s.t the coordinates given; set f `(x) =-1 and
    f ``(x)=-10.

    I think d=30 and c=-1.
    So I need to solve for a,b.

    I didn't think solving for x would be this tricky? What have I missed?
    if you are trying to solve for x simply solve for y=0. you will get three values of x in terms of a,b,c and d. refer to theory of equations . solve the following equations obtained by the theory:
    x1+x2+x3=-b/a------------(1)
    x1.x2+x2.x3+x3.x1=c/a----------(2)
    x1.x2.x3=-d/a------------------(3)
    where x1,x2,x3 are the 3 roots.also given that d=30 from the given point.
    Last edited by Pulock2009; March 24th 2010 at 01:57 AM. Reason: forgot to mention something
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  3. #3
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    Hi,
    I am still unsure about how to apply the theory of equations to this problem. Any further information would be appreciated.
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  4. #4
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    Quote Originally Posted by Neverquit View Post
    Hello,
    I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
    It cuts the y-axis at (0,30). The gradient at this point is -1.
    The gradient where x=1, is -17 and the f ``(x) at this point is -10.

    I s.t the coordinates given; set f `(x) =-1 and
    f ``(x)=-10.

    I think d=30 and c=-1.
    So I need to solve for a,b.

    I didn't think solving for x would be this tricky? What have I missed?
     <br />
f(x) = ax^3+bx^2+cx+d<br />

    f(0) = 30 ... d = 30

     <br />
f'(x) = 3ax^2 + 2bx + c<br />

    f'(0) = -1 ... c = -1

    f'(1) = -17 ... \textcolor{red}{3a + 2b -1 = -17}

     <br />
f''(x) = 6ax + 2b<br />

    f''(1) = -10 ... \textcolor{red}{6a+2b = -10}

    solve the two equation system for a and b
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  5. #5
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    I understand the a, b, c, d (a=2,b=-11,c=-1,d=30) part. It's applying the theory of polynomials equations to solve for the roots. I haven't any experience with this type of problem.
    Any information is appreciated.
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  6. #6
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    Quote Originally Posted by Neverquit View Post
    I understand the a, b, c, d (a=2,b=-11,c=-1,d=30) part. It's applying the theory of polynomials equations to solve for the roots. I haven't any experience with this type of problem.
    Any information is appreciated.
    The Rational Roots Test
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  7. #7
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    Theory of equations solution?

    Hi..
    these are the numbers for a=2, b=-11, c=-1, d=30. st these into equation
    y=ax^3+bx^2+cx+d gives:

    y=2x^3-11bx^2-x+30

    If I do the Rational Roots test then I get
    +/- 1,2,3,5,10,15,30,1/2,3/2,5/2,15/2

    I know the roots are -3/2,5,2 but what does the solution look like with the theory of equations method solving for x1, x2, x3?
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