Find all of the solutions of the equation `z^(4)=81` Not sure how to work this out, would appreciate any help

1 Answer | Add Yours

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`z^4=81`

To solve, set one side equal to zero. To do so, subtract both sides by 81.

`z^4-81=81-81`

`z^4-81=0`

Then, factor left side.

`(z^2-9)(z^2+9)=0`

Factor z^2-9 further.

`(z-3)(z+3)(z^2+9)=0`

Then, set each factor equal to zero and solve for the values of z.

For the first factor,

`z-3=0`

`z-3+3=0+3`

`z=3`

For the second factor,

`z+3=0`

`z+3-3=0-3`

`z=-3`

And for the last factor,

`z^2+9=0`

`z^2+9-9=0-9`

`z^2=-9`

Then, take the square root of both sides.

`sqrt(z^2)=+-sqrt(-9)`

`z=+-sqrt(-9)`

Since the number inside the square root is negative, the other values of z are imaginary number.

`z=+-3i`

Hence, the solutions of the given equation are x = -3 , 3 , -3i and 3i.

We’ve answered 318,982 questions. We can answer yours, too.

Ask a question