• May 8th 2009, 02:29 PM
Phire
YouTube - Precalculus: The Essentials that Students Seem to Forget

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

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

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:

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

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

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

Even using the binomial theorem and expanding, it would probably reintroduce more strange fractional exponents. Am I right?
• May 8th 2009, 05:11 PM
Shyam
Quote:

Originally Posted by Phire
YouTube - Precalculus: The Essentials that Students Seem to Forget

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

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

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:

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

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

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

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

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

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

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

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

It becomes

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

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

and solve further.