How do you find all zeros with multiplicities of #f(x)=x^4-9x^2+14#?

1 Answer
KillerBunny
May 16, 2018

#x = -sqrt(2)#, multiplicity #1#
#x = sqrt(2)#, multiplicity #1#
#x = -sqrt(7)#, multiplicity #1#
#x = sqrt(7)#, multiplicity #1#

Explanation:

Substitute #t = x^2# (which implies #t^2=x^4#) to get

#t^2-9t+14=0#

The solutions of this equation are #t=2# and #t=7#. Remember that #t=x^2#, so you have

#t = 2 \implies x^2=2 \implies x=\pmsqrt(2)#
#t = 7 \implies x^2=7 \implies x=\pmsqrt(7)#

which are the four solutions.

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