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anonymous

  • 5 years ago

What is the solution for w: w^4-18w^2-2=0

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  1. anonymous
    • 5 years ago
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    http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.332342.html

  2. anonymous
    • 5 years ago
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    yeah, it looks like that one would be best done with subbing in x for w^2, and then using the quadratic on it, then subbing back, and you might get something pretty ugly

  3. anonymous
    • 5 years ago
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    wolfram alpha should be able to solve it too but i would just use it to compare your results

  4. anonymous
    • 5 years ago
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    I look at this and it is like Greek to me..I mean I kinda get it, this is where I get confused...

  5. anonymous
    • 5 years ago
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    Alright you take the w^2's and replace them with x's. So you will get x^2-18x-2=0. Now that you have it in the quadratic form which is ax^2+bx+c=0. You can determine the values of a, b, and c. Thus a = 1, b =-18, and c =-2. With this in mind you apply the quardratic formula which is...\[-b \pm \sqrt{b ^{2}-4ac}\div 2a\] you simply will now plug in the values of a,b,c that was listed above. You will get 2 answers.. one using the negative -b- and one using the positive -b+. And your done!

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