I'm having trouble moving terms from one side of an equation to another, particularly when I'm trying to make an equation equal to 0.
For example:
8= x^1-2x
Can someone explain o me why
-x^2+2x+8=0 is WRONG and why 0=x^2-2x-8 is RIGHT?
Thanks
I'm having trouble moving terms from one side of an equation to another, particularly when I'm trying to make an equation equal to 0.
For example:
8= x^1-2x
Can someone explain o me why
-x^2+2x+8=0 is WRONG and why 0=x^2-2x-8 is RIGHT?
Thanks
Hello
When you rearrange an equation if you follow the 'rules' then the equality will remain true.
I assume you are rearranging quadratic equations to find the roots. Equating them with zero and then factoring to find the values of x that make the expression equal to zero. On a graph theses values are the x - intercepts.
I think when finding the roots of a quadratic equation it is usual to keep the $\displaystyle x^2$ term positive, factoring seems easier that way.
Both of your equations are correct rearrangements of the origianl equation. They are the same, both being equal to zero. Although they have the same roots, x - intercepts, the two quadratic expressions are different.
$\displaystyle -x^2+2x+8 = -(x^2-2x-8)$
$\displaystyle y = -(x^2-2x-8)$
Is a reflection of the graph,
$\displaystyle y= x^2 -2x -8$
In the x axis.
See the attached graph. Thought this might help, but perhaps not. Please see Mark's post below for clarification.
Hi Ting:
I'm not sure of your intended meaning when you typed the adjective "different", so I would like to clarify.
There is a difference between the terms "equation" and "function".
There is no symbol y in the original post.
Both of the equations in the original post are mathematically equivalent.
If we replace the zeros with the symbol y, then we are expressing a functional relationship.
If we further state that y is the dependent variable in this functional relationship, then we can write:
y = f(x).
After these assumptions, you are correct in pointing out that the graph of f(x) is a reflection of the graph of -f(x).
Again, there is nothing wrong with either of the equations in the original post, and I'm still curious to know why the original poster thinks that there is.
Cheers,
~ Mark
Hello mark.
Thank you for your clarification.
The two equation are not different, both being equal to 0, but the two quadratic expressions are different.
My post is misleading, incorrect use of terminolgy, and I will edit it accordingly.
Best wishes
Ting