F(x) = 16x4 – 81 = 0 select all of the zeroes of the function.
3/2
-3/2
3/2i
-3/2i

Here we want to find all the zeros of the function f(x) = 16*x^4 - 81.

The zeros are:

x = 3/2x = -3/2x = (3/2)*ix = (-3/2)*i

First, for a given function f(x), we define the zeros as the values of x such that:

f(x) = 0.

Then we must solve:

f(x) = 16*x^4 - 81 = 0

Solving this for x leads to:

16*x^4 = 81

x^4 =  81/16

x^2 = ±√(81/16) = 9/4

x = ±√±(9/4) = ±3/2 and ±(3/2)*i

So there are 4 zeros, and these are:

x = 3/2x = -3/2x = (3/2)*ix = (-3/2)*i

Where the two complex zeros come from evaluating the second square root on -9/4.

If you want to learn more, you can read.

link

The way to figure this one out is move the 81 over to the other side so you have the equation 16x^4 = 81.  Since 81 is not divisible by 16 you have to think of another way to simplify.  16 is a perfect square, x^4 is a perfect square, and 81 is a perfect square.  So take the square root of both sides to begin the simplification.  4x^2 = 9. Divide both sides by 4 to get x^2 = 9/4. 9 and 4 are both perfect squares, so undo the square on the x by taking the square root of both sides: x = +/- 3/2.

Step-by-step explanation:

the equation is:

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