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Math Help - no solution

  1. #1
    Super Member dhiab's Avatar
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    no solution

    solve in R:

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  2. #2
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    Re: no solution

    Hey dhlab.

    What have you tried?

    Hint: Try using the rational roots theorem to see if you can get a possibility of a rational root first off.
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  3. #3
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    Re: no solution

    Write it as \displaystyle \begin{align*} X^4 - 4X^3 + 8X^2 - 4X + 3 = 0 \end{align*} with \displaystyle \begin{align*} X = x^2 \end{align*}, which can be solved using the Quartic Formula. Then solve for \displaystyle \begin{align*} x \end{align*}...
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  4. #4
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    Re: no solution

    Could you please guide further as to how do we use quadratic formula for a bi-quadratic equation.
    thanks

    Quote Originally Posted by Prove It View Post
    Write it as \displaystyle \begin{align*} X^4 - 4X^3 + 8X^2 - 4X + 3 = 0 \end{align*} with \displaystyle \begin{align*} X = x^2 \end{align*}, which can be solved using the Quartic Formula. Then solve for \displaystyle \begin{align*} x \end{align*}...
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  5. #5
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    Re: no solution

    I didn't say Quadratic Formula, I said Quartic Formula.

    I suggest you read this:
    Attached Files Attached Files
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  6. #6
    Super Member dhiab's Avatar
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    Re: no solution

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  7. #7
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    Re: no solution

    What do you mean by saying " (1- x^2)^4\ne 0? It certainly is equal to 0 or some values of x.

    (You are correct that 2x^4+ 2 is never 0 and that is all you need to show that the sum is not 0.)
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  8. #8
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    Re: no solution

    by using x = y = B/4 we get
    ( y+1)^4 - 4 ( y+1)^3 + 8(y+1)^2 + 3 = 0
    y^4+2y^2 +8y+8=0
    (y^2+1)^2 = 8y -7
    How do we proceed from hereonward
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