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Math Help - Another HW Question Polynomial and Radical Equations

  1. #1
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    Another HW Question Polynomial and Radical Equations

    Use factoring to solve the equation for real values of the variables

    z^3/2 - 2^1/2 = 0

    I think i'm making it harder then it realy is please help
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  2. #2
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    Hi again civiliam,

    We have z^{\frac{3}{2}}-2^{\frac{1}{2}}=0 . Now this leads to z^{\frac{3}{2}}=2^{\frac{1}{2}} . Notice that to get z we need to raise z^{\frac{3}{2}} to the power of \frac{2}{3} since we know that (a^b)^c=a^{bc} . Hence raising both sides of the equation to the power \frac{2}{3} yields z=2^{\left(\frac{1}{2}\right)\left(\frac{2}{3}\rig  ht)}=2^{\frac{1}{3}} .

    Hope this helps.
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  3. #3
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    This is what I would do:

    z^{3/2} - 2^{1/2} = 0

    z^{3/2} = 2^{1/2}

    (z^{3/2})^{2/3} = (2^{1/2})^{2/3}

    z = 2^{1/3}
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  4. #4
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    sorry why 2/3?
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  5. #5
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    Quote Originally Posted by civiliam View Post
    sorry why 2/3?
    Because 2/3 is the reciprocal of 3/2. When you raise z^{3/2} to the 2/3 power, the exponent on z becomes 1.
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  6. #6
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    ahh I c thank you much!
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  7. #7
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    so i wrote the problem out wrong would I still work it out the same way if the 2^1/2 was z^1/2?

    z^3/2 - z^1/2 = 0
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  8. #8
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    Now you can factor out z^{1/2}.

    Beginning with z^{3/2} - z^{1/2} = 0:

    z^{1/2} \cdot (z - 1) = 0

    z = 0 or 1.

    So the problem changes significantly.
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  9. #9
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    wow that was simple odd how minor changes to a problem can throw me off so easily thanks again
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