How do you find the zeros of #y=4x^4-11x^2-3#?

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Answer 1 · 2016-09-22T23:51:06

#x=+-1/2i#

#x=+-sqrt(3)#

#4x^4-11x^2-3=0# can be factored

#(4x^2+1)(x^2-3)=0#

So we have

#4x^2+1=0# or #x^2-3=0#

Proceeding solve each factor for #x#

#4x^2=-1#

#x^2=-1/4#

#x=+-sqrt(1/4)i#

#x=+-1/2i#

#x^2=3#

#x=+-sqrt(3)#