Is There an Algebra Mistake in this Derivation?

  • Thread starter ManyNames
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Also - in response to your summary, the last line is not right. The original problem had v+1 in it, not v. It's not a big deal, but in this type of problem you might want to be careful about small details.In summary, by dividing both sides by (v+1) and carefully following the algebraic steps, the correct solution for M is M = \sqrt{\frac{k-v}{v}} which can also be written as M = \sqrt{\frac{k-v}{v+1}}. The original mistake of distributing M with (v+12) leads to an incorrect expression and should be avoided.
  • #1
ManyNames
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And so here is the derivation, and i am quickly checking here to see if the algebra is fine.

[tex]k=M(v+1^2)[/tex]
which is the same as
[tex]Mv+M^2=k[/tex]
divide by v from both sides
[tex]M+M^2=\frac{k}{v}[/tex]
then take away M from both sides
[tex]M^2=\frac{k}{v} -1[/tex]
take the square root
[tex]M=\sqrt{\frac{k}{v} -1}[/tex]
which is the same as
[tex]M=\sqrt{\frac{k-v}{v}}[/tex]
so by rearranging
[tex]M \sqrt{v}=\sqrt{k-v}[/tex]

See any mistakes i should be aware of?
 
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  • #2
by the way, equation 3 seems to be coming out wrong. Even though i have changed it, the right hand side should read, k/v
 
  • #3
The error is when distributing M with (v + 12). 12 is 1 so this gives M(v + 1) = Mv + M. So, solving for M does not, in this case, require distributing but rather dividing both sides by (v + 1).
 
  • #4
chrisk said:
The error is when distributing M with (v + 12). 12 is 1 so this gives M(v + 1) = Mv + M. So, solving for M does not, in this case, require distributing but rather dividing both sides by (v + 1).


I thought this, thanks... but otherwise, the rest would have been fine?
 
  • #5
In other words, i could state k=M(M + v), and that would solve it yes?
 
  • #6
I'll just assume then it would be correct. There was more.

Near enough where i last left this;
[tex]M=\sqrt{\frac{k-v}{v}}[/tex]
so
[tex]M\sqrt{v}=\sqrt{k-v}[/tex]
Removing the radical sign
[tex](M\sqrt{v})^2[/tex]
And then multiply by v would give
[tex]MMv^2[/tex]
So now divide by 2 gives
[tex]\frac{MMv^2}{2}=\frac{1}{2}(MMv^2)[/tex]
Rearranging gives, using again the notion that [tex]MMv^2=(M\sqrt{v})^2[/tex]
[tex]\frac{1}{2}Mv^2=\frac{(M\sqrt{v^2})^2}{2M}[/tex]

But it is these last steps i was unsure about. Can you notice any visible mistakes for me to correct?
 
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  • #7
anyone?
 
  • #8
I'm not sure what the objective of this question is, but your algebra is incorrect in Step 3 and step 4. What should the final answer look like? you want to find the kinetic energy of the system?

As chrisk suggested, the easy way is to divide both sides by (v+1) in first step.

Ultimately, this
[tex]
M=\sqrt{\frac{k-v}{v}}
[/tex]
is wrong. Check your steps carefully.
 
  • #9
Could you explain how it is wrong, carefully, because i thought it was right...
 
  • #10
The reason why i never devided, by the way, both sides by (v+1), because i wanted to isolate it first.
 
  • #11
ManyNames said:
The reason why i never devided, by the way, both sides by (v+1), because i wanted to isolate it first.
Isolate what?

ManyNames said:
Could you explain how it is wrong, carefully, because i thought it was right...

[tex]
Mv+M^2=k
[/tex]
Dividing both sides by v does not give you
[tex]
M+M^2=\frac{k}{v}
[/tex]
and taking M away from both sides won't give you
[tex]
M^2=\frac{k}{v} -1
[/tex]
 
  • #12
Dividing by both sides, i thought would have removed the v in the Mv part. This is wrong i take it?

Also, taking M from both sides, i was going to write on the right hand side k/v-M, but since M was gained from the 1 part in (v+1), i thought i could bring the value back... but i can't?
 
  • #13
Starting from
[tex]Mv+M^2=k[/tex]
Dividing by v from both sides:
[tex]\frac{Mv}{v}+\frac{M^2}{v}=\frac{k}{v}[/tex]
which simplifies to
[tex]M+\frac{M^2}{v}=\frac{k}{v}[/tex]

And starting from
[tex]M+M^2=\frac{k}{v}[/tex]
Taking M away from both sides:
[tex]M+M^2-M=\frac{k}{v}-M[/tex]
Which simplifies to
[tex]M^2=\frac{k}{v}-M[/tex]
M on the right hand side cannot be simplified to 1. The 1 part in (v+1) is lost when you re-wrote the first equation and expanded it.

Make sure you fill in the middle step when you are doing the algebra (like I did above). This makes mistakes easier to spot.
 
  • #14
Oh i see now. I apologize, i wasn't thinking straight. Yes, that makes much more sense. Is the rest right?
 
  • #15
I have a question as well. In equation 3, the v vanishes from the left hand side. How did this happen?
 
  • #16
ManyNames said:
I have a question as well. In equation 3, the v vanishes from the left hand side. How did this happen?

I'm assuming you're referring to the original post:
[tex]
M+M^2=\frac{k}{v}
[/tex]

That was your mistake. If you did the correct math, the v does not vanish from left hand side as I showed in my last post:
[tex]
M+\frac{M^2}{v}=\frac{k}{v}
[/tex]
 
  • #17
The error that Crisk found means that there is no M^2 and therefore no square root in the problem! You have to go back to the beginning and avoid that mistake which totally changes the expression you have to work with for the rest of the problem.

It might be an idea to check the original question - that 1 squared is rather an odd thing to put in a question. Why not just plain "1" ?
 

FAQ: Is There an Algebra Mistake in this Derivation?

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Algebra is a branch of mathematics that uses symbols and letters to represent quantities and the relationships between them.

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The basic operations in algebra are addition, subtraction, multiplication, and division. These operations are used to manipulate and solve equations.

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To solve an algebra equation, you need to isolate the variable by performing the same operation on both sides of the equation. This will help you find the value of the variable.

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