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Find all the zeroes of the equation
-3x^4+27x^2+1200=0
To find the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0, we can apply the quadratic formula. However, we first need to factor out a common factor of -3 from all terms:
-3(x^4 - 9x^2 - 400) = 0
Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -9, and c = -400. We can use the quadratic formula to find the zeroes:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
x = (-(-9) ± √((-9)^2 - 4(1)(-400))) / (2(1))
Simplifying further:
x = (9 ± √(81 + 1600)) / 2
x = (9 ± √(1681)) / 2
x = (9 ± 41) / 2
Now we have two possible solutions for x:
1. When x = (9 + 41) / 2:
x = 50 / 2
x = 25
2. When x = (9 - 41) / 2:
x = -32 / 2
x = -16
Therefore, the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0 are x = 25 and x = -16.