LCD of two factors?

Question

(x)/(x^2-81) and (3x)/(x^2+18x+81).........i need the LCD again. please explain. and if you want to seperatly give me tips n finding LCD's that would be great too.

well. if you know it then tell me how. not just gloat that you know how. i suck at math. thats just me. but oddly enough i am a wiz at astronomy and space (considering i have never taking a class on it).

Answer 1

The least common denominator (LCD) of two fractions is the lowest denominator (or you could think about it as simplest denominator) that both expressions have in common.

Step 1 in figuring this out for algebraic expressions is to factor both denominators, just like you would do for number denominators.
x^2 - 81 = (x+9)(x-9)
x^2 + 18x + 81 = (x+9)(x+9)

Then, you need to determine the smallest combination that includes all these factors for the LCD.
-- If you just used all of the factors for the LCD (x+9)(x+9)(x+9)(x-9), then you have a larger number/more than you need.
-- The smallest common denominator only has the two factors from the second expression and the (x-9) from the first expression. The third (x+9) is a duplicate in both expressions.
--> The LCD is (x+9)(x+9)(x-9)

Answer 2

First look at X^2 - 81, it is the difference of two squares so it's factors are: (x+9)(x-9)...

for x^2 + 18x + 81, it is a perfect square (x + 9)^2

now you compare (x+9)(x-9) and (x+9)(x+9).. and pick out the first factor (x+9) from each...
this will leave you with two terms; (x-9) from the first one.. and (x+9) from the second one...

this gives you 3 factors for your LCD = (x+9)(x+9)(x-9)...

working with your first function; (x)/[(x+9)(x-9)] you can easily see that all it needs on the bottom to be your LCD is another (x+9) so you multiply the whole thing by (x+9)/(x+9).. (which is the same as multiplying by 1) to get [(x)(x+9)]/[(x+9)(x+9)(x-9)]

working with your second function (3x)/[(x+9)(x+9) you can see that it is missing the (x-9) to be the same as your LCD.. so multiply by (x-9)/(x-9) to get: [(3x)(x-9)]/[(x+9)(x+9)(x-9)]

You don't say if you are adding or subtracting as your reason to find an LCD so I'll leave it at that point...

it is easier with numbers.. say you have two numbers.. 6 and 10 and you need the LCD...
factors of 6 = 2 * 3
factors of 10 = 2 * 5
take the 2 from each since it is common and then you need the 3 and the 5.. so you have 2*3*5 as the LCM (the only factor you leave out is one of the 2's because it is common to both and you only need it one time)

If you had 12 = 2*2*3
and 20 = 2*2*5
then you pick out the 2*2 (which is common to both) and you still need the 3 and the 5.. so you get 2*2*3*5 for the LCM between 12 and 20...

this is related to LCD if you had numbers like 1/12 and 1/20 then the LCD would be 2*2*3*5...

ok.. I'm tired and I didn't explain that well.. so I'll just leave it at that.

Answer 3

(x)/(x^2-81) and (3x)/(x^2+18x+81)

x^2-81=(x+9)(x-9)

x^2+18x+81=(x+9)^2

LCD=(x-9)(x+9)^2

Answer 4

Dude, I'm in 5th grade, and I know that! Try going, to like, Cosmeo.com and sign up.