Can someone show me how this is solved algebraically

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Homework Statement


I need to show how the first equation gives the second equation algebraically
Two relativistic energy equations stacked vertically. Top: E = m0c^2 divided by the square root of [1 - ((u' + v)/(1 + vu'/c^2))^2 / c^2]. Bottom: E = m0c^2(1 + vu'/c^2) divided by the square root of [(1 - v^2/c^2)(1 - u'^2/c^2)].



Homework Equations



I've tried multiple methods such as rationalizing the denominator with no avail.

The Attempt at a Solution



Any help will be greatly appreciated,

Thanks.
 
chris_0101 said:

Homework Statement


I need to show how the first equation gives the second equation algebraically
View attachment 51672


Homework Equations



I've tried multiple methods such as rationalizing the denominator with no avail.

The Attempt at a Solution



Any help will be greatly appreciated,

Thanks.

Multiply out the numerator of that mess over c^2. Then find a quick common divisor. Then use the rules :

[itex]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/itex]

and

[itex]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/itex]
 
I'm not sure what you mean by "multiply out the numerator of that mess over c^2"?
 
chris_0101 said:
I'm not sure what you mean by "multiply out the numerator of that mess over c^2"?

The quantity [itex](\frac{u'+v}{1+vu'/c^2})^2[/itex]
 
yea, but how does one multiply out the numerator?
do you mean like this:
(a/b)/c = (a/b)*(1/c)
 
chris_0101 said:
yea, but how does one multiply out the numerator?
do you mean like this:
(a/b)/c = (a/b)*(1/c)

...

[itex](\frac{u'+v}{1+vu'/c^2})^2 = (\frac{u'+v}{1+vu'/c^2})(\frac{u'+v}{1+vu'/c^2})[/itex]

Now some simple bedmas and you're done...
 
This is what I got:

(u'^2 + 2u'v + v^2)/((u'^2 v^2)/c^4) + (2((u'v)/c^2) +1

Is this right?