How do you find the zeroes of #f(x) = x^4= -7x^2 -144#?

1 Answer
Jul 20, 2015

I found:
#x_1=4#
#x_2=-4#
#x_3=3i#
#x_4=-3i#

Explanation:

I start supposing that the second #=# sign is not necessary, so:
#f(x)=x^4-7x^2-144#
You can find the zeroes (#x# values that makes your function equal to zero) by setting your function equal to zero and get:
#x^4-7x^2-144=0#

set #x^2=u# so you get:
#u^2-7u-144=0#
Using the Quadratic Formula you get:
#u_(1,2)=(7+-sqrt(49-4(-144)))/2=(7+-25)/2#
so you get 2 solutions:
#u_1=16#
#u_2=-9#
But #x^2=u# so #x=+-sqrt(u)#

You get 4 zeroes for your function (2 Real and 2 Immaginary):
#x_1=sqrt(16)=4#
#x_2=-sqrt(16)=-4#
#x_3=sqrt(-9)=3i#
#x_4=-sqrt(-9)=-3i#