Find all the zeroes of the equation. show all steps –3x^4+ 27x^2 + 1200 = 0
To find the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0, we can substitute y = x^2 to turn this into a quadratic equation in y.
-3y^2 + 27y + 1200 = 0
Now, we can find the zeroes of this quadratic equation by using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -3, b = 27, and c = 1200. Plugging these values into the quadratic formula:
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 for y are:
y = (-27 + 123) / -6 = 96 / -6 = -16
y = (-27 - 123) / -6 = -150 / -6 = 25
Now, remember that y = x^2, so:
1) x^2 = -16
x = ±√(-16) = ±4i
2) x^2 = 25
x = ±√25 = ±5
Therefore, the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0 are x = -5, 5, -4i, 4i.