How do you solve the quadratic using the quadratic formula given #x^4+16x^2-225=0# over the set of complex numbers?

Question

2076 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-08T07:10:24

#x in {-5i, 5i, -3, 3}#

Using the quadratic formula, along with that #sqrt(-x) = isqrt(x)# for any #x in RR#:

#x^4+16x^2-225=0#

#=> (x^2)^2+16(x^2)-225 = 0#

#=> x^2 = (-16+-sqrt(16^2-4(1)(-225)))/(2(1))#

#=(-16+-34)/2#

#=-8+-17#

#=> x^2 = -25 or x^2 = 9#

#=> x = +-sqrt(-25) or x = +-sqrt(9)#

#=> x = +-5i or x = +-3#

#:. x in {-5i, 5i, -3, 3}#