# Factoring Help

1. Sep 17, 2009

### AC5FFw

While a problem, this is more for me to learn to help my kid with her own homework. I used to be great at this sort of thing but for some reason I just can't get my head around this one.

-125 = 39.1t - 4.9t^2

I need to find the value for t. It should be approx 10.4. I would just like to see the steps to get there.

I know that I should re-write this as : 4.9t^2 - 39.1t - 125 = 0 But after this I am stuck...

Thx!

2. Sep 17, 2009

### CFDFEAGURU

Are you familiar with the quadratic formula? That is how the problem is solved.

Thanks
Matt

3. Sep 17, 2009

### AC5FFw

Don't think so. Guess I could google it.

4. Sep 17, 2009

### CFDFEAGURU

Check it out.

Thanks
Matt

5. Sep 17, 2009

### HallsofIvy

Staff Emeritus
Very few quadratic equations can be solved by "factoring". All can be solved by "completing the square" or the "quadratic formula". That doesn't look like it will have rational roots and so cannot be "factored" in the usual way.

6. Sep 17, 2009

### AC5FFw

I don't think I was ever taught this. So, when I look at an equation like the one above I try to see how to simplify it further. WoW what Fun!

7. Sep 17, 2009

### AC5FFw

Plugged this formula into the quadratic equation and got my 10.4 as one of the results.

With the +/- option, I take it I have to figure out both then plug them back into the equation. One works and the other does not. So I know that "t" has to equal 10.4

8. Sep 17, 2009

### Staff: Mentor

For your equation, both of the values you found are solutions. Since your quadratic equation comes from what appears to be a physics problem, a negative value for t might not be applicable, since that would represent a time before the object was released.

9. Sep 17, 2009

### LumenPlacidum

To illustrate the fact that both values you get are solutions, you can use a graph. A quadratic equation such as the one you gave can be graphed. For example, you can rewrite your equation by putting all the terms on one side of the equal sign:

4.9 x^2 - 39.1 x - 125 = 0

Then, instead of having '0' on the right side of the equation, you can put in a dependent variable, 'y'. Now, solutions to your original equation are points on a graph where y=0, or what are called 'roots' of the equation. Due to the shape of such a graph, it is apparent that you frequently have *two* answers that work, like so:

Where the red graph of your quadratic cross the black x-axis (where y is equal to zero), you have the solutions to your problem.

10. Sep 17, 2009

### AC5FFw

Lumen...

Your graph/pic did not come up.. but that's more of a issue with my server here.. All sorts of restrictions. I will check this out when I get home this evening...

Thank you all! I'm definatly a lot smarter on this now than I was yesterday! LOL
Where were all these great helpful tools when I was in school! If I would have had the resources that are available now... I would have enjoyed my time in school a lot more, and probably would have gone much further in my education as well.. :)

Please.. dont say it... I know, it's "Never Too Late" :D