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Really? I can't do Algebra? Little Help Here

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    How do we get from the first line to the second?

    Picture1-28.png

    I can see that the denominator vanishes along with the factor of [itex]\pi\omega^5m^2Y[/itex] in the numerator.

    What the hell is the next step? How are they getting rid of the [itex](k-m*\omega^2)^2[/itex] term?

    Can some make a suggestion as to how to simplify this? should I expand the [itex](k-m*\omega^2)^2[/itex] ?

    Is there some trick or am I mentally challenged this week?

    Casey
     
  2. jcsd
  3. Apr 7, 2009 #2
    Alrighty... If I expand out the numerator I get something like this:

    [itex]k^2-2m\omega^2+m^2\omega^2+c^2\omega^2-2c\omega^2=0[/itex]

    Which is a quadratic (where c is the independent var) whose coefficients A,B, &C are:

    A=[itex]\omega^2[/itex]

    B=[itex]-2\omega^2[/itex]

    C=[itex]k^2-2m\omega^2+m^2\omega^2[/itex]

    So I have:

    [tex]c=\frac{2\omega^2\pm\sqrt{4\omega^4-4\omega^2(k^2-2m\omega^2+m^2\omega^2)}}{2\omega^2}[/tex]

    Now where do I go? I see that there 4 coming out of the radicand as a 2 so I get:

    [tex]c=\frac{\omega^2\pm\sqrt{\omega^4-\omega^2(k^2-2m\omega^2+m^2\omega^2)}}{\omega^2}[/tex]
     
  4. Apr 7, 2009 #3
    Would it be silly to square both sides to get rid of that radicand? Eff it...I'm doing it! Who's with me?!?!?

    Don't worry...I'll just talk myself through this one.

    Now we have:

    [tex]c^2=\frac{\omega^4\pm\omega^4-\omega^2(k^2-2m\omega^2+m^2\omega^2)}{\omega^4}[/tex]

    Now for these pesky omega^2 's

    Hold on now...I am getting a little sloppy...
     
    Last edited: Apr 7, 2009
  5. Apr 7, 2009 #4
    This is stupid.

    Anyone see a better way?
     
  6. Apr 7, 2009 #5
    This isn't right. I expanded out the numerator and set it equal to zero. I ended up with the expression:
    [tex]k^2-2kmw^2+m^2w^4-c^2w^2=0[/tex]
    I then found their answer from this. Is there any reason why c must be positive? Is this some physical problem?

    Yes, it is silly to do such a thing. It is a cardinal rule of algebra that you cannot distribute an exponent over addition! You should also give people time to post as well.
     
  7. Apr 7, 2009 #6
    Ok, first of all, I am assume that [tex]\omega[/tex], m, and y are constants.

    After we dispose of the denominator, we have the numerator equal to zero.

    Notice in the first line of the handwritten image you posted, the expression [tex]\pi\omega^{5}m^{2}y[/tex] is a common factor of both terms on either side of the negative sign. If we factor thatou,we get:

    [tex]\left(\pi\omega^{5}m^{2}y\right)\left[\left(k-m\omega^{2}\right)^{2} + c^{2}\omega^{2} - 2c^{2}\omega^{2}\right] = 0[/tex]

    Now, I can divide both sides by [tex]\pi\omega^{5}m^{2}y[/tex] since it is one big constant:

    [tex]\left(k-m\omega^{2}\right)^{2} + c^{2}\omega^{2} - 2c^{2}\omega^{2} = 0[/tex]

    Combine like terms:

    [tex]\left(k-m\omega^{2}\right)^{2} - c^{2}\omega^{2} = 0[/tex]

    [tex]\left(k-m\omega^{2}\right)^{2} = c^{2}\omega^{2}[/tex]

    Take the square root of both sides:

    [tex]k-m\omega^{2} = c\omega[/tex]

    [tex]\frac{k-m\omega^{2}}{\omega} = c[/tex]
     
  8. Apr 7, 2009 #7
    Holy Sh** ... i completely missed that factor of "c" in the last term.....thanks meiso...
    I knew I was special-ed this week..
     
  9. Apr 7, 2009 #8
    Yes. But come to find out, it's a mental one too.

    Damn Cardinal rules! You are very right!

    You are probably right, but I like to be proactive even if it's not productive. :smile:
     
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