How do you find the number of complex zeros for the function #f(x)=x^4-9x^2+18#?

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Answer 1 · 2018-03-08T13:18:19

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We can make the variable change #t=x^2#. With this change we obtain

#t^2-9t+18=f(t)#. If we are lookin form zeros od #f#, then

#t^2-9t+18=0#. Apliying knowed formula

#t=(-b+-sqrt(b^2-4ac))/(2a)=(9+-sqrt(81-72))/2=(9+-9)/2#

Both solutions for #t# are: #t=0# and #t=9#

Undoing the change of variable

#t=0=x^2#, so #x=0# (double root)
#t=9=x^2#, so #x=+-3#

Our initial equation has no complex roots (or zeros), but we can express them in complex form (with their imaginary parts zero)

#0+0i# double
#3+0i#
#-3+0i#