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Math Help - Finding the real solutions

  1. #1
    Super Member
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    Finding the real solutions

    Hi

    Can someone help me how to solve the following:

    1) Find the real solutions of the cubic equation: 9x^3 + 8x +6

    This is what i have done:

    let x = x

    9x^2 +8+6 = 0

    9x^2 = -14

    x^2=\frac{-14}{9}

    x=1.2472

    2) Find the inequality corresponding to the graph:


    P.S
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Paymemoney View Post
    Hi

    Can someone help me how to solve the following:

    1) Find the real solutions of the cubic equation: 9x^3 + 8x +6

    This is what i have done:

    let x = x , what do u mean ?

    9x^2 +8+6 = 0 , how did the x^3 become x^2

    9x^2 = -14

    x^2=\frac{-14}{9} , you can't have negative values in the square root

    x=1.2472 , this is wrong


    hi

    this can't be solved algebraically (Perhaps it can , but i don see how). The fastest way to solve it is by using a calculator OR you can also approximate the roots using the newton-raphson method .
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  3. #3
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    Quote Originally Posted by mathaddict View Post

    hi

    this can't be solved algebraically (Perhaps it can , but i don see how). The fastest way to solve it is by using a calculator OR you can also approximate the roots using the newton-raphson method .

    ok thanks,

    also the second question forgot to add that there were 4 answers to choose from:

    y>0.272x+0.428-\frac{2}{x}

    y>-0.419x-0.360

    y>-1.291x-1.45

    y>-0.521x+0.603-\frac{1}{x}

    y>1.97x-0.142-\frac{1}{x^2}
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  4. #4
    MHF Contributor

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    It should be obvious that the lower boundary of that region is a straight line and so its equation is linear. That excludes all except the second and third. In the picture it looks like when x= 0, y is between -.25 and -.50. Which of the given options is that?
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