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Math Help - Quadratic Substitution.

  1. #1
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    Quadratic Substitution.

    Solve Using substitution.

    x^{3/2}+8x^{1/2}+16x^{-1/2}=0

    Let A= x^{1/2}

    A^2+8A+\frac{16}{A}=0

    A^3+8A^2+16=0

    A^2(A+8)=-16

    I have no clue if i even did this correctly...Help me please.
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  2. #2
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    Solve Using substitution.

    x^{3/2}+8x^{1/2}+16x^{-1/2}=0

    Let A= x^{1/2}

    A^2+8A+\frac{16}{A}=0
    This should be

    A^3+8A+\frac{16}{A}=0

    Now multiply through by A.
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  3. #3
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    Re: Quadratic Substitution.

    So even if the the first factor is not ^2 of the second term, A substitution is still possible?

    A^4+8A^2+16=0

    A^2(A^2+8)=-16

    A=\pm\sqrt{-16}
    Or
    A=\pm\sqrt{-24}
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  4. #4
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    So even if the the first factor is not ^2 of the second term, A substitution is still possible?

    A^4+8A^2+16=0

    A^2(A^2+8)=-16

    A=\pm\sqrt{-16}
    Or
    A=\pm\sqrt{-24}
    A^4+8A^2+16=0

    Now A^2 = B thus

    B^2+8B+16=0

    (B+4)^2=0 , back substitute twice.
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  5. #5
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    Re: Quadratic Substitution.

    A=\pm\sqrt{-4}
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    A=\pm\sqrt{-4}
    Yes, now you know that \displaystyle x^{\frac{1}{2}} = \pm \sqrt{-4}, what is \displaystyle x?
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  7. #7
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    Re: Quadratic Substitution.

    Not sure since there is already a \pm on one side, how do I square both sides?
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    Not sure since there is already a \pm on one side, how do I square both sides?
    What happens when you square a positive number? What happens when you square a negative number?
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  9. #9
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    Re: Quadratic Substitution.

    x=\pm4

    x=\pm-4
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  10. #10
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    x=\pm4

    x=\pm-4
    Not even close, sorry.

    \displaystyle \left(\sqrt{-4}\right)^2 = \sqrt{-4} \times \sqrt{-4} = -4.

    \displaystyle \left(-\sqrt{-4}\right)^2 = \left(-\sqrt{-4}\right)\left(-\sqrt{-4}\right) = \sqrt{-4}.

    Therefore \displaystyle \left(\pm \sqrt{-4}\right)^2 = -4.
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