• Support PF! Buy your school textbooks, materials and every day products Here!

Problem with quadratics (dividing by x)

  • Thread starter david18
  • Start date
  • #1
49
0
Hi, could someone please tell me the circumstances under which you can divide a quadratic by the variable?

e.g. x^2 - (15/2)x = 0. I know I'd usually have to go through the whole factorisation/formula process and get more than one answer for x, but if I know that x is positive am I able to simply divide it by x to get x - 15/2 = 0?

The reason I ask is because this is what I do in my projectile motion questions. Whenever I get a question in which the vertical displacement is 0 (e.g. ball being kicked up then falling back to ground) I end up with the equation 0 = ut + 1/2 at^2 (from the equation of motion s = ut + 1/2 at^2). I always divide this by t and get the right answer, but is this what I'm meant to be doing?

Any help would be greatly appreciated,
David
 

Answers and Replies

  • #2
454
0
when you try to divide the general quadratic ax^2 + bx + c = 0 by x, what is the term that can't be divided by x most of the time? for what value(s) of the constants can this term also be divided by x?

when you divide by x, you really factor the quadratic, and one of the factors is x, so the quadratic still has 2 solutions, one of them is x = 0
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
249
Whenever I get a question in which the vertical displacement is 0 …
Hi David! :smile:

If the vertical displacement is 0, then t = 0 will always be a solution, won't it (because after zero time it will always have moved zero distance! :biggrin: )?

So you will always be able to divide the quadratic by t! :smile:
 

Related Threads for: Problem with quadratics (dividing by x)

Replies
13
Views
6K
Replies
19
Views
858
  • Last Post
Replies
11
Views
1K
Replies
16
Views
890
Replies
7
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
612
Replies
1
Views
1K
Top