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Thread: YouTube lesson about precalculus wrong?

  1. #1
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    YouTube lesson about precalculus wrong?

    YouTube - Precalculus: The Essentials that Students Seem to Forget

    It starts at 5:29. Basically, what he is saying is that:

    $\displaystyle x^{\frac{1}{3}} + x^{\frac{1}{2}} - 7 = 0$

    He's says that you can take the LCM of the fractional exponents (6), and multiply both sides of the equation by x^6 to remove the fractions. But, if you do that, wouldn't this happen:

    $\displaystyle x^{\frac{6}{1}} * x^{\frac{1}{3}} = x^{\frac{18}{3} + \frac{1}{3}} = x^{\frac{19}{3}}$

    Which doesn't help at all.. And even if you tried something like this:

    $\displaystyle (x^{\frac{1}{3}} + x^{\frac{1}{2}})^6 = 7^6$

    Even using the binomial theorem and expanding, it would probably reintroduce more strange fractional exponents. Am I right?
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  2. #2
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    Quote Originally Posted by Phire View Post
    YouTube - Precalculus: The Essentials that Students Seem to Forget

    It starts at 5:29. Basically, what he is saying is that:

    $\displaystyle x^{\frac{1}{3}} + x^{\frac{1}{2}} - 7 = 0$

    He's says that you can take the LCM of the fractional exponents (6), and multiply both sides of the equation by x^6 to remove the fractions. But, if you do that, wouldn't this happen:

    $\displaystyle x^{\frac{6}{1}} * x^{\frac{1}{3}} = x^{\frac{18}{3} + \frac{1}{3}} = x^{\frac{19}{3}}$

    Which doesn't help at all.. And even if you tried something like this:

    $\displaystyle (x^{\frac{1}{3}} + x^{\frac{1}{2}})^6 = 7^6$

    Even using the binomial theorem and expanding, it would probably reintroduce more strange fractional exponents. Am I right?
    Yes, you are RIGHT. He is teaching the wrong method.

    Actually it should be

    $\displaystyle x^{\frac{1}{3}} + x^{\frac{1}{2}} - 7 = 0$

    $\displaystyle x^{\frac{2}{6}} + x^{\frac{3}{6}} - 7 = 0$

    $\displaystyle (x^{\frac{1}{6}})^2 + (x^{\frac{1}{6}})^3 - 7 = 0$

    Then, substitute $\displaystyle x^{\frac{1}{6}} = t$

    It becomes

    $\displaystyle t^2+t^3-7=0$

    $\displaystyle t^3+t^2-7=0$

    and solve further.
    Last edited by Shyam; May 8th 2009 at 05:33 PM.
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