LCD

sunshine_200202

New member
Joined
Sep 11, 2005
Messages
1
Do I have the right LCD for this?

3/ x+1 + x+5/ x^2-1 + 3/ 1-x

x+1= x+1
x^2-1 = (x+1)(x-1)
1-x= 1-x

LCD= (x+1) (x-1), (x+1)

Thank you!

~Diana
 
Well...
If you change the +3 to -3 and the 1-x to x-1 the LCD would be (x-1)(x+1) or x²-1.
BTW Use more ()s. You use some but 4/2+6 doesn't = 1/2, it =8. You MUST say 4/(2+6) if you want 1/2
 
You have posted as follows:

. . . . .(3/x) + 1 + x + (5/x<sup>2</sup>) - 1 + (3/1) - x

I think you mean the following:

. . . . .3/(x + 1) + (x + 5)/(x<sup>2</sup> - 1) + 3/(1 - x)

If so, and if the instructions are to find the Least Common Denominator, then note:

. . . . .x<sup>2</sup> - 1 = (x + 1)(x - 1)
. . . . .1 - x = -1(x - 1)

Since no denominator has more than one copy of the "x + 1" factor, the LCD would not contain more than one copy of it, either.

Eliz.
 
I carefully avoided Eliz's factoring approach 'cause I have a problem. If the denominator (1-x) were Q(x-1) then Q would be part of the LCD. I couldn't see why, just because Q= -1, it isn't. Must be an LCD rule I've forgotten.
 
The "-1" can go anywhere:

. . . . .1/[-1(x - 1)] = -1/(x - 1) = - [1/(x - 1)]

Eliz.
 
That still leaves me with the Q question. The Q could go to the numerator as 1/Q but deep in my heart I still feel it part of the LCD.
-----------------
Gene
 
Well, yes, technically, you do make a good point. But the usual expectation of the textbooks is that "the '1' can go anywhere". But if the factor were anything other than "1", yes, you'd have to account for that in the LCD. You're right about that.

Eliz.
 
Top