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Find all the zeroes of the equation.
−3x4+27x2+1200=0
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, we can let y = x^2. This way, we can rewrite the equation as a quadratic equation in terms of y.
-3y^2 + 27y + 1200 = 0
Now, we can solve this quadratic equation for y using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = -3, b = 27, and c = 1200.
y = (-27 ± √(27^2 - 4(-3)(1200))) / 2(-3)
y = (-27 ± √(729 + 14400)) / -6
y = (-27 ± √15129) / -6
y = (-27 ± 123) / -6
Therefore, the two possible values of y are:
y = (123 - 27) / -6 = -96/6 = -16
y = (-123 - 27) / -6 = -150/6 = -25
Now that we have the values of y, we can substitute back to find the values of x (since y = x^2):
For y = -16:
x^2 = -16
x = √(-16) = ±4i
For y = -25:
x^2 = -25
x = √(-25) = ±5i
Therefore, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are x = ±4i and x = ±5i.