Quote Originally Posted by OneMileCrash View Post
Alright, I gave in and looked at my solutions manual.

They're going from:

$\displaystyle |\frac{-5(x-4)}{ (\sqrt{x-3} )(\sqrt{2x-1}-\sqrt{7x-21} )} |$


$\displaystyle |x-4| \times \frac{5}}{ (\sqrt{x-3)(\sqrt{2x-1}-\sqrt{7x-21} )} $

Why is it legal for me to just take the ABV out of that whole division operation? The abv would have previously been of the denominator too, but now it isn't, shouldn't that matter?
if we let a = x-4, and b the ugly denominator, we have |-5a/b| = |-5a|/|b| = |-5||a|/|b| = 5|a|/|b|.

as noted in prevous posts, we have to have x > 3, so one factor of the denominator > 0.

is the solution from your manual really the difference √(2x-1) - √(7x-21)? it seems that this should be a sum

(because as x-->4, √(2x-1) - √(7x-21) --> √7 - √7 = 0, which means the reciprocal would be unbounded).

again, it should be clear that if x > 3, x > 1, so √(2x-1) > 0, and √(7x - 21) > 0, so their sum is positive.

and i just realized something: in all my previous posts, i had this term as √(7x+21) (this stems from it being wrong in your post #8).

however, the arguments are still valid, because x > 3.5 > 3, so 7x-21 > 0, which is all my argument requires.

one thing about proofs of this nature is: watch your signs! i count at least 4 sign errors that have occured in this thread, 3 of which completely derail

your proof. anyway, continuing....

since both factors in your denominator are positive (by suitably keeping x "close enough to 4"), the absolute value delimiter can be removed.

that is, in our "shorthand" 5|a|/|b| = 5|a|/b, since b > 0.