Re: Please Help: an algebraic problem

On Sat, 9 Apr 2005, Chris Quickfall wrote:

> Can anyone help to erradicate the denominator: (2r-t) from the
> expression;
>
> ( r^2 - (3/2)tr + t^2 ) / ( 2r - t )
>
First get 1/2 out of the way.

(2r^2 - 3tr + 2t^2) / 2(2r - t)

Now t = 2r isn't a soluton, so division will leave remainder.
Thus you're stuck with the ulgy fact 2r - t ain't going away.

What you want to do is follow the example of the WhiteLie House
and change the facts to your convient liking.

(2r^2 - 3rt + t^2) / 2(2r - t)
(2r - t)(r - t) / 2(2r - t)
or
(-2r^2 - 3rt + 2t^2) / 2(2r - t)
(2r - t)(-r - 2t) / 2(2r - t)
that way you can get what you want even tho you shouldn't.

> I have been trying to factorise the top into a form of (2r-t), so that
> i can cancel the denominator, but cannot quite get there. This is what
> i've done:
>
> (2r-t)((1/2)r - t) = r^2 - (5/2)tr + t^2
> (2r-t)(-(1/2)r + t) = -r^2 + (5/2)tr - t^2
>
> (2r-t)((1/2)r + t) = r^2 + (3/2)tr - t^2
> (2r-t)(-(1/2)r - t) = -r^2 - (3/2)tr + t^2
>
> As you can see i am close but either the sign is always wrong, or my
> fractions are always out. Is there anyone who can solve? I'm stumped.
>
.

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