Question

Help how do I find the roots!

x(2x+3)(x^2+9)(3x^2-1)=0

Answer 1

To find the roots of the equation:
x(2x+3)(x^2+9)(3x^2-1)=0
we need to set each factor equal to zero and solve for x.
x = 0
2x+3 = 0 -> 2x = -3 -> x = -3/2
x^2+9 = 0 -> x^2 = -9 -> x = ±3i (where i is the imaginary unit)
3x^2-1 = 0 -> 3x^2 = 1 -> x^2 = 1/3 -> x = ±(1/√3)
Therefore, the roots of the equation are:
x = 0, -3/2, 3i, -3i, 1/√3, and -1/√3.