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Luv2Teach
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.
facundo3141592

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

The zeros are:

  • x = 3/2
  • x = -3/2
  • x = (3/2)*i
  • x = (-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 = 3/2
  • x = -3/2
  • x = (3/2)*i
  • x = (-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.

https://brainly.com/question/8815381

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