How do you find the complex roots of #4m^4+17m^2+4=0#?

1 Answer
Nov 26, 2016

Answer:

The roots are:

#m = +-1/2i" "# and #" "m = +-2i#

Explanation:

#4m^4+17m^2+4 = 0#

Treating this as a quadratic in #m^2#, we can use an AC method to factor it into quadratics, then use the difference of squares identity with Complex coefficients.

First find a pair of factors of #AC = 4*4 = 16# with sum #B=17#.

The pair #16, 1# works.

Use this pair to split the middle term, then factor by grouping:

#4m^4+17m^2+4 = 4m^4+16m^2+m^2+4#

#color(white)(4m^4+17m^2+4) = (4m^4+16m^2)+(m^2+4)#

#color(white)(4m^4+17m^2+4) = 4m^2(m^2+4)+1(m^2+4)#

#color(white)(4m^4+17m^2+4) = (4m^2+1)(m^2+4)#

#color(white)(4m^4+17m^2+4) = ((2m)^2-i^2)(m^2-(2i)^2)#

#color(white)(4m^4+17m^2+4) = (2m-i)(2m+i)(m-2i)(m+2i)#

Hence zeros:

#m = +-1/2i" "# and #" "m = +-2i#