Solving this algebra equation for z.?

Question

Can somebody please help me solve this algebra equation?

(5z^2 + 6z +12)^2 = 0

Thank you very much

Answer 1

The square is zero, so you can just remove it (take the sqrt of both sides). Now solve by using quadratic formula.

a = 5
b = 6
c = 12

x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

x = [-6 +/- sqrt(6^2 - 4(5)(12)) ] / 2(5)

x = -3/5 +/- sqrt(36 - 240) / 10

x = -3/5 +/- sqrt(-204) / 10
x = -3/5 +/- 1/5 sqrt(-51)

x = -3/5 +/- 1/5 sqrt(51) i

So you have two imaginary roots

Answer 2

The last squaring is irrelevant. The only way (...)^2 can be 0
is for (...) to be 0.

so solve
5z^2 + 6z + 12 = 0

The quadratic formula

z = (-b + or - sqrt(b^2 - 4ac))/2a
with a = 5, b = 6, c = 12

will give a negative value for
b^2 - 4ac,

which means you get complex roots.
Do it yourself, but I think you get
z = -0.6 + 0.2*sqrt(51)*i
or z = -0.6 - 0.2*sqrt(51)*i

Answer 3

z=4

Answer 4

Write it out like this:

(5z^2 + 6z + 12)(5z^2 + 6z + 12),

which extends to:

(5z^2)(5z^2 + 6z + 12) + (6z)(5z^2 + 6z + 12) + (12)(5z^2 + 6z + 12)

Answer 5

(5z^2 + 6z +12)^2 = 0 this is really a lot simpler than it looks. just take the sqrt of both sides & get:
(5z^2 + 6z +12) = 0
z=(-6+/-sqrt(36-240))/10
z=(-6+/-sqrt(-204))/10=-3/5+/-1/5sqrt(-51)
z=-.6-.2sqrt(51) i
z=-.6+.2sqrt(51) i

Answer 6

quadratic formula

z=-6/10+-[sqrt(36-240)]/10
= -3/5 +-[sqrt(204]i/10

Answer 7

the only solutions are complex (not real numbers)

(3 + or - i*sqrt(51)) / 5

Answer 8

factor the quadratic

Answer 9

ok i dont know