How do you find all the real and complex roots of #-4x^4+3x^2+1#?

1 Answer

Answer:

#x=-1/2i# and #x=+1/2i# and #x=-1# and #x=+1#

Explanation:

Let us start from the given
#-4x^4+3x^2+1=0#
this is also
#4x^4-3x^2-1=0#
Using Factoring Method
#(4x^2+1)(x^2-1)=0#
and equating both factors to zero we have

#(4x^2+1)=0# this is the first factor

#4x^2=-1" "# divide both sides by 4

#(cancel4x^2)/cancel4=-1/4" "#

#x^2=-1/4" "#take the square root of both sides

#sqrt(x^2)=+-sqrt(-1/4)#

#x=+-1/2i#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the other factor

#(x^2-1)=0#

#x^2=1" " "#take the square root of both sides

#sqrt(x^2)=+-sqrt(1)#

#x=+-1#

Just because this is a 4th degree equation, we are expecting 4 possible roots or zeros

#x=-1/2i# and #x=+1/2i# and #x=-1# and #x=+1#

God bless...I hope the explanation is useful..