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Thread: Quadratic Substitution.

  1. #1
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    Quadratic Substitution.

    Solve Using substitution.

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

    $\displaystyle Let A= x^{1/2}$

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

    $\displaystyle A^3+8A^2+16=0$

    $\displaystyle A^2(A+8)=-16$

    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.

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

    $\displaystyle Let A= x^{1/2}$

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

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

    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?

    $\displaystyle A^4+8A^2+16=0$

    $\displaystyle A^2(A^2+8)=-16$

    $\displaystyle A=\pm\sqrt{-16}$
    $\displaystyle Or$
    $\displaystyle A=\pm\sqrt{-24}$
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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?

    $\displaystyle A^4+8A^2+16=0$

    $\displaystyle A^2(A^2+8)=-16$

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

    Now $\displaystyle A^2 = B$ thus

    $\displaystyle B^2+8B+16=0$

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

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

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

    Not sure since there is already a $\displaystyle \pm$ 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 $\displaystyle \pm$ 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.

    $\displaystyle x=\pm4$

    $\displaystyle x=\pm-4$
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    Re: Quadratic Substitution.

    Quote Originally Posted by theloser View Post
    $\displaystyle x=\pm4$

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

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

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

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