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help with solving for x

  • 01-05-2017 12:14pm
    #1
    manutd
    Registered Users, Registered Users 2 Posts: 988 ✭✭✭


    solve the following equation for x (correct to 1 decimal place)
    5x^3-35x^2+47x=0
    do you use the formula ax^3 + bx^2 + cx + d = 0


Comments

  • #2
    Bit cynical
    Registered Users, Registered Users 2 Posts: 2,338 ✭✭✭Bit cynical


    Note that in your equation, there's no constant term, i.e. no equivalent to d in the second equation. This suggests that there's an operation you can perform on the left hand of your equation to simplify things.

    Is there some common factor in each of the terms on the left hand side that can be factored out?


  • #3
    SomethingElse
    Registered Users, Registered Users 2 Posts: 1,534 ✭✭✭SomethingElse


    x=0

    Or

    5x^2-35x+47=0

    Solving:

    x=5.188, 1.812 or x=0


  • #4
    manutd
    Registered Users, Registered Users 2 Posts: 988 ✭✭✭manutd


    Bit cynical wrote: »
    Note that in your equation, there's no constant term, i.e. no equivalent to d in the second equation.  This suggests that there's an operation you can perform on the left hand of your equation to simplify things.

    Is there some common factor in each of the terms on the left hand side that can be factored out?
    x is a common factor.


  • #5
    Bit cynical
    Registered Users, Registered Users 2 Posts: 2,338 ✭✭✭Bit cynical


    manutd wrote: »
    x is a common factor.
    Well you basically have it then. Divide both sides by x and solve the quadratic. x=0 becomes an additional root to the two roots of the quadratic equation.


  • #6
    TheBody
    Registered Users, Registered Users 2 Posts: 5,629 ✭✭✭TheBody


    Bit cynical wrote: »
    Well you basically have it then. Divide both sides by x and solve the quadratic. x=0 becomes an additional root to the two roots of the quadratic equation.

    The correct procedure is to factorise and then solve.

    You can't divde both sides by x because you potentially could be dividing by zero.


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