Solve 2x^3 + 9x^2 -18x = 0 for x.
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The equation 2x^3 + 9x^2 -18x = 0 has to be solved.
2x^3 + 9x^2 -18x = 0
=> x*(2x^2 + 9x - 18) = 0
=> x*(2x^2 + 12x - 3x - 18) = 0
=> x*(2x(x + 6) - 3(x + 6)) = 0
=> x*(2x-3)(x + 6) = 0
This gives x = 0
2x - 3 = 0
=> x = 3/2
x + 6 = 0
=> x = -6
The solutions of the equation are {-6, 0, 3/2}
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First, you want to factor out anything that is common in each of the terms. In this equation, that is the x. So:
x(2x^2+9x-18)=0
Then, you separate this into two equations,
x=0 and 2x^2+9x-18=0 (because if x=0, then the whole equation will be equal to zero).
Now you need to find the remaining values of x:
1. Factor: (2x-3)(x+6)=0
2. Separate: 2x-3=0, x+6=0
3. Solve for x in both equations:
2x-3=0
2x=3
x=3/2
and
x+6=0
x=-6
Therefore, the solutions to this equation (called the "zeros" of the function because they occur when the graph crosses the x-axis) are: x=0, x=3/2, and x=6
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