To solve the equation -3x^4 + 27x^2 + 1200 = 0, we can substitute y = x^2, which gives us the equation -3y^2 + 27y + 1200 = 0.
Next, we can solve this quadratic equation for y. Using the quadratic formula, y = (-b ± √(b^2 - 4ac))/(2a), where a = -3, b = 27, and c = 1200.
y = (-(27) ± √((27)^2 - 4(-3)(1200)))/(2(-3))
= (-27 ± √(729 + 14400))/(-6)
= (-27 ± √(15129))/(-6)
= (-27 ± 123)/(-6)
Therefore, y = (-27 + 123)/(-6) or y = (-27 - 123)/(-6).
For y = (-27 + 123)/(-6), we get y = 96/(-6) = -16.
For y = (-27 - 123)/(-6), we get y = -150/(-6) = 25.
Now, we substitute y back into the equation y = x^2, so x^2 = -16 or x^2 = 25.
For x^2 = -16, there are no real solutions because the square of a real number cannot be negative.
For x^2 = 25, we take the square root of both sides, giving x = ±√(25).
Therefore, the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0 are x = -5 and x = 5.
Find all the zeroes of the equation.
–3x4+ 27x2 + 1200 = 0
1 answer
1 answer