Find all of the real and imaginary solutions of x^3+8=0

Step-by-step explanation:

Real solutions ():

However, recall that .

Therefore:

.

Thus, for :

.

Therefore, the real and nonreal solutions of are .

-2, 1 +/- i*sqrt(3)

(3 solutions)

Step-by-step explanation:

Because the left side is in the form a^3+b^3, we can factor it as (a+b)(a^2-ab+b^2) (good formula to remember)

So we have

(x+2)(x^2-2x+4)

We can see one value already

x+2 = 0

x = -2

now we solve for

x^2-2x+4 = 0

We can use the quadratic formula, and see that the determinant is negative (2^2-4*1*4 < 0)

Therefore we have two imaginary solutions

(-(-2) +/- sqrt(2^2-4*1*4))/)2*1) = (2 +/- sqrt(-12))/2 = (2 +/- 2sqrt(-3))/2 = 1+/- sqrt(-3) = 1 +/- i*sqrt(3)

at least two

step-by-step explanation:

given the equation is 8+9z=-6, the first step would be to subtract the 8 from both sides:

8+9z-8=-6-8

since 8 minus 8 = 0, this would cancel out the 8. and since -6 minus 8 equals -14, the equation would look like this

9z = -14