How do you solve the quadratic using the quadratic formula given #x^4+13x^2+36=0# over the set of complex numbers?

1 Answer
Feb 15, 2017

Answer:

#+-2 i and +-3 i#

Explanation:

#x^4 + 13 x^2 + 36=0#

assuming #y = x^2#,

#(x^2)^2 + 13 x^2 + 6 =y^2+13y+36=0#

using #y = (-b +- sqrt(b^2-4ac))/(2a)#, where #a=1, b=13 and c=36#

#y = (-13+-sqrt(13^2-4(1)(36)))/(2(1))#

#y = (-13+-sqrt(169-144))/2#

#y = (-13+-sqrt(25))/2#

#y = (-13+-5)/2#

#y = (-13+5)/2=-4#, #y = (-13-5)/2=-9#

when #y =-4, x^2 =-4#

#x = sqrt(-4) = sqrt(4*(-1))= sqrt4*sqrt(-1)=+-2i#

when #y =-9, x^2 =-9#

#x = sqrt(-9) = sqrt(9*(-1))= sqrt9*sqrt(-1)=+-3i#