1. ## x-intercept

I cannot remember how to work this problem but here is what I think I am suppose to do.
The problem states "Solve and give the solutions x=

5x+x(x-2)=0
do I mulitply
5x+x^2 -2x = 0
3x + x^2 = 0
3x - 3x + x^2 = 0 - 3x
This is the point were I feel I am on the wrong track can you help?

2. Originally Posted by Brama Bull
I cannot remember how to work this problem but here is what I think I am suppose to do.
The problem states "Solve and give the solutions x=

5x+x(x-2)=0
do I mulitply
5x+x^2 -2x = 0
3x + x^2 = 0
3x - 3x + x^2 = 0 - 3x
This is the point were I feel I am on the wrong track can you help?
At the point where you have $\displaystyle 3x + x^2 = 0$ you have it in a form that is easily factorable, therefore easily solvable.

Factor out the common x term from the left side:

$\displaystyle x(3 + x) = 0$

Then solve for x.

3. ## x-intercept

the problem states find the x- intercept
f(x)=5x+x(x-2).
this is what I have done
5x+x^2-2x
x^2+3x
x(x+3)
now were do I go from here to find the x-intercept?

4. Originally Posted by Brama Bull
the problem states find the x- intercept
f(x)=5x+x(x-2).
this is what I have done
5x+x^2-2x
x^2+3x
x(x+3)
now were do I go from here to find the x-intercept?
Are you certain the problem does not ask for the y intercept? In the equation you wrote, which is quadratic, you have two real roots. They're normally not referred to as "x intercepts" in this context.

5. Set $\displaystyle x(x+3)=0$, so your x int are 0. - 3

6. Yes it asks for the x- intercept; it can be in theses forms;(type an ordered pair, or simply your answer, type an exact answer using radicals as needed, express complex numbers in terms if i, Use a comma to seperate answers as needed)

but I have double checked and it says the x-intercepts? I already have the solutions as (0,-3)

7. It may be reading (0,-3) as an ordered pair. Try this (0,0) and (-3,0)