Answer:
3/2, -3/2, (if it is asking for imaginary solutions as well then 3i/2 and -3i/2)
Step-by-step explanation:
All zeros of the function is,
x = 3/2
x = -3/2
x = (3/2)i
x = (-3/2)i
Every complex number may be represented in the form a + bi, where a and b are real numbers. A complex number is an element of a number system that extends the real numbers with a specific element labeled I sometimes known as the imaginary unit, and satisfying the equation i² = 1.
Given
Here we want to find all the zeros of the function f(x) = 16x⁴ - 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) = 16x⁴ - 81 = 0
Solving this for x leads to:
16x⁴ = 81
x⁴ = 81/16
x² = ±√(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.
To learn more about complex zeros refer to:
https://brainly.com/question/11442072
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