Is this the correct way to factor this equation?

[ManyNames]

Homework Statement

Needing to factor the relevant equation.

Homework Equations

(pM)= ½ (F²vt/c)+ ½ (F²vt/v)

Square both sides:

(pM)²= ½ (F²vt/c)² + ½ (F²vt/v)²

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

 

[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.

 

[Joskoplas]

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

 

[ManyNames]

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.

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.

 

[ManyNames]

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

How would i do that exactly?

Thanks by the way.

I ask because i am bit confused about my operations.

 

[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.

 

[ManyNames]

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.

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

 

[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).

 

[ManyNames]

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).

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?

 

[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)

 

[ManyNames]

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)

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

 

[ManyNames]

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)

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 you 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 ;)

 

[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:

<br /> 2 p M = F^2 t ( \frac{v}{c} - 1 )<br />

 

[ManyNames]

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:

<br /> 2 p M = F^2 t ( \frac{v}{c} - 1 )<br />

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?

 

[Joskoplas]

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

 

[Joskoplas]

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