Z4=289 there should be 4 answers total 2 whole numbers and either 2square root or imaginary number answers

There are 4 solutions and they are:

where i = sqrt(-1) and also "sqrt" is shorthand for "square root"

Work Shown:

Let x = z^2. Square both sides to get x^2 = (z^2)^2 = z^(2*2) = z^4

In short, x^2 = z^4, so we can replace the 'z^4' with 'x^2' to get this new equation x^2 = 289

Apply the square root to both sides

x^2 = 289

sqrt(x^2) = sqrt(289)

sqrt(x^2) = sqrt(17^2)

|x| = 17

x = 17 or x = -17

For each equation above, replace x with z^2 and solve that equation for z

x = 17

z^2 = 17

sqrt(z^2) = sqrt(17)

|z| = sqrt(17)

z = sqrt(17) or z = -sqrt(17)

which are two solutions out of the four total

The other two solutions are found by plugging x = z^2 into x = -17 and solving for z

x = -17

z^2 = -17

sqrt(z^2) = sqrt(-17)

|z| = sqrt(-1*17)

|z| = sqrt(-1)*sqrt(17)

|z| = i*sqrt(17)

z = i*sqrt(17) or z = -i*sqrt(17)

which are the other two solutions

u dumb boi

step-by-step explanation:

the answer will be the last one