First the short way. The apparent solutions from graphing the equations are x=0 and x=±√5
So the product of the non-zero solutions is -5
I wanted to confirm this so I raised both sides to the 15th power, then simplified the result to a 22nd degree polynomial with all even terms.
Using the substitution y=x² brings us to a 10th degree polynomial (too big to type out here)
I was able to use synthetic division to show y=5 is a zero of the polynomial so x²=5 and so indeed x=±√5
I was also able to find the some non-real solutions y=x²=-4 so x= ±2i and also an irrational y=x²≈.0024990669
Posted by Jer
on 2014-02-23 21:35:56