How do you find the number of complex, real and rational roots of #2x^4-18x^2+5=0#?

1 Answer
Nov 8, 2016

Answer:

Find the zeros.

Explanation:

First substitute #a# for #x^2#.
#2a^2-18a+5=0#

Next find what #a# is using the quadratic formula:
#a=\frac{18\pm\sqrt((-18)^2-4(2)(5))}{2(2)}#

Next, simplify:
#a=\frac{18\pm\sqrt(324-40)}{4}#
#a=\frac{18\pm\sqrt(284)}{4}#
#a=\frac{18\pm2\sqrt(71)}{4}#
#a=\frac{9\pm\sqrt(71)}{2}#

#a=x^2#, so:
#x^2=\frac{9\pm\sqrt(71)}{2}#
#x=\pm\sqrt(\frac{9\pm\sqrt(71)}{2})#

So the four zeros are:
#\sqrt(\frac{9-\sqrt(71)}{2}),\sqrt(\frac{9+\sqrt(71)}{2}),-\sqrt(\frac{9-\sqrt(71)}{2}),-\sqrt(\frac{9+\sqrt(71)}{2})#

#root()(71)# is less than 9, so all of the zeros are real. #sqrt(71)# is irrational, so none of the zeros will be rational.