find all the zeroes in the equation
-2x^(4)-64x^(2)+288=0
To find the zeros of the equation -2x^4 - 64x^2 + 288 = 0, we can use the quadratic formula since it is a quadratic equation in terms of x^2.
Let's substitute x^2 as a variable y to simplify the equation: -2y^2 - 64y + 288 = 0.
Now, using the quadratic formula, the solutions for y are given by:
y = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -2, b = -64, and c = 288.
Substituting these values into the quadratic formula:
y = (-(-64) ± √((-64)^2 - 4(-2)(288))) / (2*(-2))
= (64 ± √(4096 + 2304)) / (-4)
= (64 ± √6400) / (-4)
= (64 ± 80) / (-4)
Now, let's consider the positive square root:
y = (64 + 80) / (-4) = 144 / (-4) = -36
Substituting y = -36 back to x^2:
x^2 = -36
Since we cannot take the square root of a negative number for real solutions, there are no real zeros for x in the given equation.