SOLUTION: Find the values of X for which f(x)=g(x) f(x)=2x^2-4 g(x)=3x^3+18 My work so far: 2x^2-4=3x^3+18 -3x^3+2x^2-4-18=0 -3x^3+2x^2-22=0 (-x- )(3x^2+ )=0 I am confu

Algebra ->  Trigonometry-basics -> SOLUTION: Find the values of X for which f(x)=g(x) f(x)=2x^2-4 g(x)=3x^3+18 My work so far: 2x^2-4=3x^3+18 -3x^3+2x^2-4-18=0 -3x^3+2x^2-22=0 (-x- )(3x^2+ )=0 I am confu      Log On


   

Question 1027877: Find the values of X for which f(x)=g(x)
f(x)=2x^2-4 g(x)=3x^3+18

My work so far:
2x^2-4=3x^3+18
-3x^3+2x^2-4-18=0
-3x^3+2x^2-22=0
(-x- )(3x^2+ )=0 I am confused about this step and how I should continue further


Thanks for the help!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You're trying to factor a difficult equation.
You can't do it closed form easily, you need to solve it numerically.
3x%5E3-2x%5E2%2B22=0
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I graphed it to get close.
Then used Newton's method to solve for the real root.
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