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Math Help - Solving equations, but what type

  1. #1
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    Solving equations, but what type

    To solve an equation I would require first to be able to correctly identify it?

    This is a typical equation I am looking at;

    x^3 = 7x^2 + 12x = 0

    At first I thought it was a quadractic but now I am not sure because the equation ax2 + bx + c = 0 does not contain a place for the x which is multiplied by the 12.

    So if this type of equation is not a quadratic, would somebody please advise what type it is then I can have a go at solving it please.

    Thank you
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Solving equations, but what type

    This is a third degree polynomial.
    Initially, you will learn to solve these by
    1) factor by grouping
    2) rational roots theorem/synthetic division

    Of course in this case you can factor:
    x(x^2 + 7x + 12) = 0,
    then factor the quadratic the usual way.
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  3. #3
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    Re: Solving equations, but what type

    Quote Originally Posted by TheChaz View Post
    This is a third degree polynomial.
    Initially, you will learn to solve these by
    1) factor by grouping
    2) rational roots theorem/synthetic division

    Of course in this case you can factor:
    x(x^2 + 7x + 12) = 0,
    then factor the quadratic the usual way.
    Thanks I'll start learning these types of equations now. Can't understand why they are in the course really as I can't find any information in the book(s) supplied except the quadratic formulas. These formulas give two routes or solutions, but the above I quote has three solutions?

    Better get reading then.

    Thanks

    David
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  4. #4
    Super Member TheChaz's Avatar
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    Re: Solving equations, but what type

    The degree of a polynomial is intimately linked with the number of solutions.
    For our purposes, let's just say that the highest exponent is the maximum number of real solutions.

    The methods I mentioned are more advanced. If you're only learning quadratics, they can still introduce higher-degree polynomials by multiplication with a monomial.
    For instance, you can FOIL (2x + 3)(5x - 9) if needed.
    Then if you multiply this by 11x^8 and set it to zero, you have a hard-looking problem that will reduce to factoring a quadratic after removing the GCF
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  5. #5
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    Re: Solving equations, but what type

    Quote Originally Posted by TheChaz View Post
    This is a third degree polynomial.
    Initially, you will learn to solve these by
    1) factor by grouping
    2) rational roots theorem/synthetic division

    Of course in this case you can factor:
    x(x^2 + 7x + 12) = 0,
    then factor the quadratic the usual way.
    So I have this equation which I now understand to be a cubic expression. If I am right I now require to find factor pairs for 7 and 12, these can be both positive and negative integers?

    - 1, -7, 1,7. -2, 2, -3, 3, -4, 4

    How am I doing so far?
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    Re: Solving equations, but what type

    Quote Originally Posted by David Green View Post
    So I have this equation which I now understand to be a cubic expression. If I am right I now require to find factor pairs for 7 and 12, these can be both positive and negative integers?

    - 1, -7, 1,7. -2, 2, -3, 3, -4, 4

    How am I doing so far?
    The closest I have got so far is;

    x^3 + 7x^2 = 12x = 0

    x(x^2 - 4)(7x + 12x)

    x^3 + 7x^2 + 12x - 28x - 48x

    x^3 + 7x^2 + 8

    I can't be far out now, I just need to find the correct factor so that my + 8 becomes + 12
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  7. #7
    Super Member TheChaz's Avatar
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    Re: Solving equations, but what type

    Start with x^3 + 7x^2 + 12x = x(x^2 + 7x + 12) = 0

    x^2 + 7x + 12 factors into (x + 3)(x + 4), separately from the x.

    So you have x(x + 3)(x + 4) = 0
    and set each factor to zero.
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  8. #8
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    Re: Solving equations, but what type

    Quote Originally Posted by TheChaz View Post
    Start with x^3 + 7x^2 + 12x = x(x^2 + 7x + 12) = 0

    x^2 + 7x + 12 factors into (x + 3)(x + 4), separately from the x.

    So you have x(x + 3)(x + 4) = 0
    and set each factor to zero.
    In the above I worked that out to show; x^2 + 7x + 12. Is there something missing because I thought that when the brackets were multiplied out that the solution should be the same as the original equation?

    Also when I am looking for the factors, what do you mean by saying, set each factor to zero?

    Please advise
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  9. #9
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    Re: Solving equations, but what type

    Quote Originally Posted by David Green View Post
    In the above I worked that out to show; x^2 + 7x + 12. Is there something missing because I thought that when the brackets were multiplied out that the solution should be the same as the original equation?

    Also when I am looking for the factors, what do you mean by saying, set each factor to zero?

    Please advise
    Just had a senior moment, I was working out the brackets incorrectly?

    x(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12

    x(x^2 + 7x + 12) = x^3 + 7x^2 + 12x

    It seems that I have only found two solutions though?

    (x+3) and (x+4)

    How do I go about finding the third solution?

    Sorry Chaz you did say previously that the highest power is x^3, which I read afterwards in a maths book, but I am still unclear how to present it as three solutions in this example?
    Last edited by David Green; September 8th 2011 at 02:02 PM. Reason: Misunderstanding of information given
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  10. #10
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    Re: Solving equations, but what type

    You have now written "x^3 = 7x^2 + 12x = 0", in your original post, "x^3 + 7x^2 = 12x = 0", in post number 6, and The Chaz wrote "x^3 + 7x^2 + 12x = 0" in post number 7. which is it?


    If it is, indeed, x^3 + 7x^2 + 12x = x(x^2+ 7x+ 12)= x(x+ 3)(x+ 4)= 0, then the "zero property" gives you all three roots: the product of x(x+ 3)(x+ 4) is 0 only if at least one of the three factors is 0: x= 0, or x+ 3= 0, or x+ 4= 0.

    You also seem to be very confused about exactly what you are trying to do. You say "It seems that I have only found two solutions though? (x+3) and (x+4)". Those are NOT "solutions". The three solutions you want are the three numbers that satisfy the equation. If x+ 3= 0 what is x? If x+ 4= 0, what is x? Now what was the third factor, above?
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  11. #11
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    Re: Solving equations, but what type

    Quote Originally Posted by HallsofIvy View Post
    You have now written "x^3 = 7x^2 + 12x = 0", in your original post, "x^3 + 7x^2 = 12x = 0", in post number 6, and The Chaz wrote "x^3 + 7x^2 + 12x = 0" in post number 7. which is it?


    If it is, indeed, x^3 + 7x^2 + 12x = x(x^2+ 7x+ 12)= x(x+ 3)(x+ 4)= 0, then the "zero property" gives you all three roots: the product of x(x+ 3)(x+ 4) is 0 only if at least one of the three factors is 0: x= 0, or x+ 3= 0, or x+ 4= 0.

    You also seem to be very confused about exactly what you are trying to do. You say "It seems that I have only found two solutions though? (x+3) and (x+4)". Those are NOT "solutions". The three solutions you want are the three numbers that satisfy the equation. If x+ 3= 0 what is x? If x+ 4= 0, what is x? Now what was the third factor, above?
    Please accept my applogies, the original equation should be;

    x^3 + 7x^2 + 12x = 0

    x(x + 3)(x + 4) = 0

    x^2 + 4x + 3x + 12 = 0

    x(x^2 + 7x + 12) = 0

    x^3 + 7x^2 + 12x = 0

    x = 0
    x + 3 = 0
    x + 4 = 0

    to get that;

    - 3 + 3 = 0
    - 4 + 4 = 0

    the three roots therefore must be;

    x = 0
    (-3 + 3) = 0
    (-4 + 4) = 0

    Please advise if I am now understanding this correctly, also remember I am learning from first principles.

    Thanks

    David
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  12. #12
    MHF Contributor Siron's Avatar
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    Re: Solving equations, but what type

    You seems confused about the solutions. You solve it correctly because the third degree equation can be factored as:
    x(x+3)(x+4)=0

    So you have the solutions:
    x=0
    x+3=0 \Leftrightarrow x=-3
    x+4=0 \Leftrightarrow x=-4

    Take a look at the post of HallsofIvy. A solution always looks like: x=...
    But (-3+3)=0 doesn't make sense as a solutions of an equation.
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