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Math Help - Radicals Part 4

  1. #1
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    Radicals Part 4

    I really appreciate this help guys, it means a lot to me. Heres another problem I can't figure out.


    x = 21 - squareroot(x-9)
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    Quote Originally Posted by Trentt View Post
    I really appreciate this help guys, it means a lot to me. Heres another problem I can't figure out.


    x = 21 - squareroot(x-9)
    We have,
    (x^2-21)=-sqrt(x^2-9)
    Let y=x^2
    Thus,
    y-21=-sqrt(y-9)
    Square,
    (y-21)^2=y-9

    This is a quadradic, the solution is trivial.
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  3. #3
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    That didn't help, but thanks anyways
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  4. #4
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    Quote Originally Posted by Trentt View Post
    That didn't help, but thanks anyways
    What is the problem.
    If you are asking more complicated equations with radicals than you certainly know quadradic equations. I am not going to solve those, you are expected to do that.
    Did you not understand how I arrived at that final equation.
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  5. #5
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    I'll just wait for Soroban, he makes it easy to understand. Thanks.
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  6. #6
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    Quote Originally Posted by Trentt View Post
    I really appreciate this help guys, it means a lot to me. Heres another problem I can't figure out.


    x = 21 - squareroot(x-9)
    rewrite as:

    x^2-21 = -sqrt(x^2-9)

    now square both sides:

    (x^2 - 21)^2 = x^2 - 9

    expand the square on the left:

    x^4 - 42 x^2 + 441 = x^2 - 9.

    Now bring everything over to the left hand side:

    x^4 - 43 x^2 + 450 =0

    which is a quadratic in x^2, which may be solved using the quadratic formula
    to give: x^2 = 25, or x^2 = 18.

    So we have x=+/5, and x=+/-sqrt(18) are the roots, but we have to
    substitute these back into the original equation because the squaring
    may have introduced spurious solutions.

    Doing this we find that the roots x=+/- 5 do not satisfy the original equation,
    while x=sqrt(18) and x=-sqrt(18) both do.

    Hence the solutions to the original equation are:

    x=sqrt(18) and x=-sqrt(18).

    RonL
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