# How can these roots be real?

• Sep 20th 2011, 06:06 AM
moriman
How can these roots be real?
I ran this equation

$112u^8+56u^4 v^4+3v^8=48u^8+24u^4 v^4+3v^8$

solving for $v$

through http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#v1=3v^8%2B24u^4+v^4%2B48u^8%3D3v^8%2B56u ^4+v^4%2B112u^8&v2=u

and am confused about it's results. It says that 2 of the values for $u$ are real. Here's one of them (sorry, I couldn't get the indices small with tex so am just using ordinary text

v=(-1)^(1/4) * 2^(1/4) * u

I don't understand how (-1)^(1/4) is real. I thought it was the same as $\sqrt{i}$

Forgot to add, I've also done this by hand and have $2u^4=-v^4$ which I think comes to the same thing.
• Sep 20th 2011, 11:54 AM
Opalg
Re: How can these roots be real?
Quote:

Originally Posted by moriman
I ran this equation

$112u^8+56u^4 v^4+3v^8=48u^8+24u^4 v^4+3v^8$

solving for $v$

through http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#v1=3v^8%2B24u^4+v^4%2B48u^8%3D3v^8%2B56u ^4+v^4%2B112u^8&v2=u

and am confused about it's results. It says that 2 of the values for $u$ are real. Here's one of them (sorry, I couldn't get the indices small with tex so am just using ordinary text

v=(-1)^(1/4) * 2^(1/4) * u

I don't understand how (-1)^(1/4) is real. I thought it was the same as $\sqrt{i}$

Forgot to add, I've also done this by hand and have $2u^4=-v^4$ which I think comes to the same thing.

The only real solution for u is u=0, in which case v can be anything. The other solutions are all complex, as you correctly say. Moral: Don't always trust free software.

To get indices to display correctly in TeX, use braces not parentheses (curly brackets rather than round ones, in other words). So for example [TEX]v=(-1)^{1/4} * 2^{1/4} * u[/TEX] yields $v=(-1)^{1/4} * 2^{1/4} * u.$
• Sep 20th 2011, 12:30 PM
moriman
Re: How can these roots be real?
Thanks for the confirmation and for the help on the indices