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Math Help - Solving a Quartic

  1. #1
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    Solving a Quartic

    Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!
    Thanks in advance
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  2. #2
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    Quote Originally Posted by theleontan View Post
    Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!
    Thanks in advance
    note that the quartic will factor ...

    x^4 - 2x^2 + 1 = (x^2 - 1)^2
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  3. #3
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    If you were not able to recognise what Skeeter posted straight away, notice that if you let \displaystyle X = x^2, then your equation becomes

    \displaystyle X^2 - 2X + 1 = 0

    which is an easy quadratic to solve. Once you have \displaystyle X you can find \displaystyle x.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    If you were not able to recognise what Skeeter posted straight away, notice that if you let \displaystyle X = x^2, then your equation becomes

    \displaystyle X^2 - 2X + 1 = 0

    which is an easy quadratic to solve. Once you have \displaystyle X you can find \displaystyle x.
    I still dont get how you went from my equation to what you have up there ^.
    Could you explain please?
    Thanks
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  5. #5
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    Quote Originally Posted by theleontan View Post
    Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real.
    Let's try again...
    let u = x^2 ; then your equation becomes: u^2 - 2u + 1 = 0
    Are you saying you can't follow that?
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  6. #6
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    Quote Originally Posted by theleontan View Post
    I still dont get how you went from my equation to what you have up there ^.
    Could you explain please?
    Thanks
    Let's look at your original equation:

    x^4 - 2x^2 +1 = 0

    note that x^4=(x^2)^2

    This means that your original equation can be rewritten as

    (x^2)^2- 2(x^2) +1 = 0

    What Prove It did was to substitute x^2 for X.
    If you substitute x^2 for X in the equation above,
    you will get this:


    (X)^2- 2(X) +1 = 0

    This equation can be solved for X with the quadratic formula, as it's a quadratic equation with X as an unknown, and as we know that X stands for x^2, this means that If we know X we can get that x=\pm \sqrt{X} , and then solve for x when we have figured out X.
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  7. #7
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    This is a longer method but the more ways you know how to do something the better! Alternatively, the factor theorem could be used.

    Subbing x = 1 into the quartic gives you 0 so x-1 is a factor. Using long division we can get a cubic answer. Using the factor theorem again on that will get you a factorisable quadratic.

    Then you will have 4 factors!
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