How do you find all zeros with multiplicities of #f(x)=2x^4-7x^2+6#?

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Answer 1 · 2017-05-28T21:09:00

See explanation.

The original equation can be transformed to a quadratic equation by a substitution: #x^2=t#

#2t^2-7t+6=0#

#Delta=(-7)^2-4*2*6=49-48=1#

#sqrt(Delta)=1#

#t_1=(7-1)/4=3/2# and #t_2=(7+1)/4=2#

Now we have to find corresponding #x# values:

#x^2=3/2#

#x_1=-sqrt(3/2)# and #x_2=sqrt(3/2)#

If we rationalize the denominators we get:

#x_1=-sqrt(6)/2# and #x_2=sqrt(6)/2#

#x^2=2#

#x_3=-sqrt(2)# and #x_4=-sqrt(2)#

Final answer:

The function has 4 real zeros:

#x_1=-sqrt(6)/2#, #x_2=sqrt(6)/2#, #x_3=-sqrt(2)#, #x_4=-sqrt(2)#