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

1 Answer
Leland Adriano Alejandro
Feb 29, 2016

#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..

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1980 views around the world
You can reuse this answer
Creative Commons License