Thread: What is the relationship between (2x^3-2x^2 = -6x^2 + 30x), and 2x^3+4x^2-30x

1. What is the relationship between (2x^3-2x^2 = -6x^2 + 30x), and 2x^3+4x^2-30x

The equations $\displaystyle 2x^3-2x^2$ and $\displaystyle -6x^2+30x$ produces intersection points at 0,0 and 3,36.

If I set them equal to each other and then solve for 0, I get the equation $\displaystyle 2x^3 + 4x^2 - 30x$ produces zeroes at 0 and 3.

So, the x's from the intersection points and the x's from the zeroes are the same. Why? The graphs look completely different, what causes them to produce the same results?

2. Originally Posted by cdbmath
The equations $\displaystyle 2x^3-2x^2$ and $\displaystyle -6x^2+30x$ produces intersection points at 0,0 and 3,36.
what you stated are not equations. there has to be an equal sign for something to be an equation.

If I set them equal to each other and then solve for 0, I get the equation $\displaystyle 2x^3 + 4x^2 - 30x$ produces zeroes at 0 and 3.
there is also a zero at -5

moreover, this is exactly the method used to find the points of intersection. why would you expect to get a different answer?

So, the x's from the intersection points and the x's from the zeroes are the same. Why? The graphs look completely different, what causes them to produce the same results?
finding the solutions to $\displaystyle 2x^3 - 2x^2 = -6x^2 + 30x$ is the same as finding the zeros of the function $\displaystyle y = 2x^3 + 4x^2 - 30x$, since you are working with exactly the SAME equation. in the latter case you are solving for $\displaystyle 2x^3 + 4x^2 - 30x = 0$, which is the same as finding the solutions to the first equation.

as for the graphs, i think you are confusing them. the graphs of the 3 expressions will of course be different. $\displaystyle y_1 = 2x^3 - 2x^2$ and $\displaystyle y_2 = -6x^2 + 30x$ are two different graphs. the latter graph $\displaystyle y_3 = 2x^3 + 4x^2 - 30x$ is also different, as it represents something else. its relation to the previous two graphs is that its zeros give the points of intersection of the two graphs

3. Originally Posted by cdbmath
The equations $\displaystyle 2x^3-2x^2$ and $\displaystyle -6x^2+30x$ produces intersection points at 0,0 and 3,36.

If I set them equal to each other and then solve for 0, I get the equation $\displaystyle 2x^3 + 4x^2 - 30x$ produces zeroes at 0 and 3.

So, the x's from the intersection points and the x's from the zeroes are the same. Why? The graphs look completely different, what causes them to produce the same results?
I think you are saying that you have the following functions:

. . . . .$\displaystyle f(x)\, =\, 2x^3\, -\, 2x^2\, =\, 2x^2(x\, -\, 1)$

. . . . .$\displaystyle g(x)\, =\, -6x^2\, +\, 30x\, =\, -6x(x\, -\, 5)$

I think you found the x-intercepts of each, and then compared the result with f(x) = g(x) or, which is the same thing, found the x-intercepts of h(x) = f(x) - g(x):

. . . . .$\displaystyle h(x)\, =\, 2x^3\, +\, 4x^2\, -\, 30x\, =\, 2x(x\, +\, 5)(x\, -\, 3)$

And now you're wondering why some of the x-intercepts were the same, but others were not.

If so, then the simple answer is that different functions may share x-intercepts, but then again, they may not.

In your case, since each of f(x) and g(x) had a factor of x, then the difference also had a factor of x, so the difference shared the x-intercept at x = 0.

However, the other factors were different, so the difference simplified differently, so the other x-intercepts were different.

If you meant something else, please reply with clarification, including the full and exact text of the exercise on which you are working. Thank you!