# Have i done this right?

1. Dec 5, 2008

### ManyNames

1. The problem statement, all variables and given/known data

Needing to factor the relevent equation.

2. Relevant equations

(pM)= ½ (F²vt/c)+ ½ (F²vt/v)

Square both sides:

(pM)²= ½ (F²vt/c)² + ½ (F²vt/v)²

3. The attempt at a solution

(pM)= ½ (F²vt/c)+ ½ (F²vt/v)

Square both sides:

(pM)²= ½ (F²vt/c)² + ½ (F²vt/v)²

Then factor and simplify

½ F²(vt/c – vt/v)= p²M²

2. Dec 5, 2008

### Joskoplas

If I square both sides of $$y = \frac{1}{2}x$$, I don't get $$y^2 = \frac{1}{2}x^2$$ ... I get $$y^2 = \frac{1}{4}x^2$$.

If I square both sides of $$y = a + b$$, I don't get $$y^2 = a^2 + b^2$$ ... I get $$y^2 = (a+b)^2 = a^2 + 2ab + b^2$$.

3. Dec 5, 2008

### Joskoplas

And might I also suggest that factoring a bit earlier might simplify the problem.

4. Dec 5, 2008

### ManyNames

In other words i have squared wrongly... which i have... hold on, because one i properly do this, i still need to factor it out correctly.

5. Dec 5, 2008

### ManyNames

How would i do that exactly?

Thanks by the way.

I ask because i am bit confused about my operations.

6. Dec 5, 2008

### Joskoplas

Well, if I had an equation of the form $$y = abx - abt$$, and wanted to solve for x, I might want to consolidate the common factors first:

$$y = ab (x - t)$$

Then move them over to the other side by dividing through by those common factors:

$$\frac{y}{ab} = x - t$$

Hmm. Only one step remains, and it can be done by adding an element to both sides.

7. Dec 5, 2008

### ManyNames

But concerning my equations... Is this relevent, because i am a terrible problem solver

8. Dec 5, 2008

### Joskoplas

What do these fellows have in common?

½ (F²vt/c)+ ½ (F²vt/v)

And I hate to ask this late in the game, but what are we solving for, here? Do we stop at factoring, or go on to solve for c or v? Because the step above is as good a run at factoring as you're likely to get. The rest is algebra (adding, subtracting, multiplying, dividing, etc through the entire equation).

9. Dec 5, 2008

### ManyNames

We are factoring out the F. And we also solve for F... if that makes sense? If it doesn't, then perhaps i need to solve for v or c?

10. Dec 5, 2008

### Joskoplas

Yep, F's easy. But try to figure out the problem yourself first. Hint:

If

$$f = \frac{abx}{c} + \frac{abx}{x}$$

then

$$f = abx(\frac{1}{c} - \frac{1}{x})$$

or equivalently (if we leave the x behind)

$$f = ab(\frac{x}{c} - 1)$$

11. Dec 5, 2008

### ManyNames

Right, o'k. I can work that out. Hold on. Cheers.

12. Dec 5, 2008

### ManyNames

pM=Fvt/c+Fvt/v

pM=Fv(t/c– 1)

Is the only pattern I see in this. Is this formula supposed to help, because I am really confused. This is the second time I have came here, funnily enough to be such confused again. But if ya could help, I’d appreciate a more direct response, because I am absolutely no good at mathematical riddles, or anything for me to solve. I don’t have a mathematical brain really ;)

13. Dec 5, 2008

### Joskoplas

Well, they're not really riddles. See, I could show you a gun and shoot a duck. Then I could hand you a gun and ask you to shoot a crow. The wrong answer in that case would be "But you didn't show me how to shoot a crow, just a duck."

Teaching by way of example is a time honored tradition. If I say I can solve y = 5x for x by dividing through by 5, I get y/5 = x. Ta daah.

But what about y = 6x? Same trick. y/6 = x

But what about y = kx, where k is some constant? Same trick. y/k = x. Ta daah.

What about a whole conga line of constants? If y = abcdefgx, hey, I can pick em off 1 by 1 like . . . well, like shooting ducks . . . and get y/(abcdefg) = x

Sometimes we answer in examples. Not riddles.

Past that? It's homework help, not solutions. I could hand over an answer, and you could repeat it, but . . .

there will be more ducks.

Anyways, your factorization left out a number of things. Try this on and see if you can work it from here. Better yet, try to figure out how I came up with this.

Given:

(pM)= ½ (F²vt/c)+ ½ (F²vt/v)

The factors can be combined to this just-about-ready-to-solve-for-F type form:

$$2 p M = F^2 t ( \frac{v}{c} - 1 )$$

14. Dec 5, 2008

### ManyNames

Simple algebraic rules otherwise, like

E=Mc^2

M=E/C^2

C^2=E/M

I know this. This is not what i was asking in the first place. But, you say the last part, and i suppose this is the proper way to factor this out for F?

15. Dec 5, 2008

### Joskoplas

Good response, now apply it and solve for F^2 as you have solved for C^2 in your example.

16. Dec 5, 2008

### Joskoplas

And you can check to see if it's a proper factorization by multiplying it out, reversing the procedure.