How do you find all the real and complex roots of #x^4 + 3x^2 + 2 = 0#?

1 Answer
Feb 6, 2016

Answer:

#x=+-i,+-isqrt2#

Explanation:

Note that #x^2+3x+2# can be factored into #(x+1)(x+2)#, so

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

So, we have the equation

#(x^2+1)(x^2+2)=0#

Set both of these equal to #0#:

#x^2+1=0#

#x^2=-1#

#x=+-i#

and

#x^2+2=0#

#x^2=-2i#

#x=+-isqrt2#

The equation has #4# complex roots and no real roots (meaning it never crosses the #x#-axis:

graph{x^4+3x^2+2 [-13.83, 14.65, -2.31, 11.93]}