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Math Help - solving an equation algebraically

  1. #1
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    solving an equation algebraically

    x^4-3=2x+1
    how can i solve it ? thanks in advance
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  2. #2
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    Quote Originally Posted by alessandromangione View Post
    x^4-3=2x+1
    how can i solve it ? thanks in advance
    x^4-2x-4=0

    I don't think you'll have much luck algebraically as both of the real roots appear to be transcendental.
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  3. #3
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    then i use the formula b^2 -square root of 4ac / 2a and i get x=0 and x=4...is it correct?
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  4. #4
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    Quote Originally Posted by alessandromangione View Post
    then i use the formula b^2 -square root of 4ac / 2a and i get x=0 and x=4...is it correct?
    That formula only works on quadratic equations and the equation in the original post is a quartic (x^4)

    Did you mean x^2-3 = 2x+1 or x^4-3 = 2x+1


    Given ax^2+bx+c=0 then the quadratic formula is x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    That formula only works on quadratic equations and the equation in the original post is a quartic (x^4)

    Did you mean x^2-3 = 2x+1 or x^4-3 = 2x+1


    Given ax^2+bx+c=0 then the quadratic formula is x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
    i mean x^4...so what formula can i use instead?
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  6. #6
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    Quote Originally Posted by alessandromangione View Post
    i mean x^4...so what formula can i use instead?
    You can't, at least not to my knowledge.

    You'd have to use technology to get an iterative solution

    Wolfram guesses too: http://www.wolframalpha.com/input/?i=x^4-3%3D2x%2B1+
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