fractional exponent within a quadratic-like equation...

[brainpolice]

OK... this was the only one I couldn't figure out. Here's what I have so far...

The original:
x^(1/3) + 3x^(1/6) - 5 = 0

3([cube root x])^(1/2) + [cube root x] - 5 = 0

Then with u-substitution (I don't think this is correct, though),

u = [cube root x]

3u^(1/2) + u - 5 = 0

...and that's where I get stuck, because I don't know how to get from 3x^(1/6) to a square power, if that makes any sense. Help please.

-Ari

 

[tkhunny]

So, you're in the same class as a prevous poster, but somehow maanged to get a different part of the lesson. Very interesting.

Try factoring. There is a reason why you have been invited to factor so many trinomials. It is time to prove it!

If it won't factor, try the Quadratic Formula (or completing the square).

Note: you may wish to try a different substitution: \(\displaystyle u = \sqrt[6]{x}\)

 

[brainpolice]

AHA! I got it now. I can't believe I didn't think of substituting the 6th root of x instead.

Thank you!

 

[tkhunny]

It is sometimes a little visually troubling. You are used to the square being larger than the un-square. When the exponent is less than unity, the roles reverse.