Your solution to the second equation is correct, and the procedure also.
For the first equation, think of it as:
x^3 + 8x^2 = 20x
x^3 + 8x^2 - 20x = 0
x(x^2 + 8x - 20) = 0
then x=0 is a solution, and you can get two more solutions from the part in parentheses in a similar way to the second equation.
x^3 + 8x^2 = 20x
How do I find the root of ^ that equation?
2x^2-10x+12 = 0
^ I found that one by...
2(x-3)(x-2)
x-3 = 0
x = 3
x-2 = 0
x =2
^ Is that the correct way to solve for that one?
3 answers
Thanks so much MathMate =)
You're welcome, and keep up the good work!