rewrite equation (rationalize denominator)

[jkristia]

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

I just started on College Algebra (online) and completed the homework for the first chapter.
There is one question which is not part of the homework, but it looks interesting so I gave it a try, but I get stuck at some point and would like some pointers of how to solve it

The questions is

Show that the equation
M = \frac {M_o} {\sqrt{1 - \frac{v^2}{c^2}}}


can be written in the form
M = \frac {M_oc \sqrt{c^2-v^2} } {c^2-v^2}

2. Relevant equations


3. The attempt at a solution

I tried to rationalize the denominator and got to

M = \frac {M_oc^2 \sqrt{1 + \frac{v^2}{c^2}} } {c^2-v^2}

But I cant see how to turn

M_oc^2 \sqrt{1 + \frac{v^2}{c^2}}

into

M_oc \sqrt{c^2-v^2}


Any help is appreciated.


Jesper

[tiny-tim]

Hi Jesper! :smile:

(have a square-root: √ :wink:)
M = M0c2 sqrt(1 + v2/c2) / (c2 - v2)

erm :redface: … where did that "+" come from? :wink:

[jkristia]

Hi Tim,
Just spend the last 1/2 hour trying to enter the equation using Latex.

argh ... found my mistake, thank you for pointint it out :)

edit - actually, that was REALLY embarrassing, next time I will triple check before asking

[Mark44]

I just started on Colage Algebra (online)

That would be College Algebra.

Collage is a word in English, but it doesn't have anything to do with education. Colage is not a word in English. College is the word you want.

[jkristia]

>>That would be College Algebra.

yes of course :)

[Slats18]

Try splitting your c^2 that is outside of the square root, into c*c and see where you can go from there =)

[jkristia]

Thanks your your help, but I found the problem as soon as tiny-tim pointed out the '+'.
This was one of my typical 'duh' mistakes. I had been doing exercises where the denominator was of form

(sqrt(a) - b)

so I had been rationalizing by multiplying with (sqrt(a) + b) in the denominator, and I did the same with this one, so that is where the '+' came from.

Of course in this case I just need to square the squareroot, and once I realized that it was straight forward.

Jesper