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    brosh4's Avatar
    brosh4 Posts: 4, Reputation: 1
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    #1

    May 6, 2012, 02:54 PM
    Solve a "quadratic type" equation
    3y4 = 28y2 - 9
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    May 7, 2012, 06:02 AM
    Quote Originally Posted by brosh4 View Post
    3y4 = 28y2 - 9
    I'm assuming you mean this:



    Start by substituting a new variable for y^2. If we let w = y^2, then this becomes



    You can rearrange to get this into a standard quadratic form, then use the quadratic formula to solve for w. There will be two soultions for w. Then finally y = sqrt(w), but be careful to check whether both positive and negative values of sqrt(w) work in the original equation.

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