Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Find the LCD of: x, X-2, and x^2-2x

OpenStudy (anonymous):

@Directrix

OpenStudy (anonymous):

What is in the denominator

OpenStudy (anonymous):

Sorry. It came out crappy. Bad. Heres how its suppose to actually be:

OpenStudy (anonymous):

\[\frac{ 3 }{ x }+\frac{ 6 }{ x-2 }=\frac{ 12 }{ x ^{2}-2x }\]

OpenStudy (anonymous):

BUT .. all i need to do is find the LCD which the LCD is x(x-2) BUT ... he said I must use the LCD chart.

OpenStudy (anonymous):

Which is like this : http://www.youtube.com/watch?v=P-GHtf9MOCk BUT ... Im not ending up with the right answer.

OpenStudy (anonymous):

The LCD is \[x ^{2} - 2x\]I don't know the LCD chart

OpenStudy (anonymous):

The LCD Chart is in the video. Its like ..... Find the LCD the way of finding the LCM .

OpenStudy (anonymous):

@Directrix

Directrix (directrix):

I don't know the term "LCD Chart" so I'm over on YouTube watching this 11-minute video to see if I recognize it.

OpenStudy (anonymous):

|dw:1383110324459:dw|

OpenStudy (anonymous):

@DocLav I did that. I multiplied x (times) x-2 ..... resulted to: x(x-2) BUT ..... using the chart .... you bring down it all because they appear once: \[x \times x-2 \times x(x-2) you will get: x ^{2} - 2x(x-2)\]

OpenStudy (anonymous):

I don't know then. Tell your teacher his LCD chart is stupid and a doctor said that. Ha

OpenStudy (anonymous):

Hahaa, yeah, it doesn't make much sense. I think the LCD Chart works best for numbers you don't have to factor.

Directrix (directrix):

It is making sense. I'm on the LCM portion of the video.

OpenStudy (anonymous):

Yeah. If there are no variables than it may work better.

OpenStudy (ranga):

No you don't multiply them all together. That is not how you find LCM. From first row you bring x down first. From second row you bring down (x-2). The third row has x and (x-2). But we already brought down x and (x-2). So we don't have to bring anything down from the third row.

Directrix (directrix):

@rbeckford134 Post the chart you made, please.

OpenStudy (anonymous):

OpenStudy (anonymous):

@Ranga ... that made a lot of sense! I think I did the chart correctly that i posted just now?

Directrix (directrix):

That is how the chart I made from the video looks. LCD = x*(x-2)

OpenStudy (ranga):

Your chart is correct for the three rows. But from each row we have to see which factors to bring down to find LCD. From the first row you bring x down. From the second row you bring (x-2) down. When we look at the third row it has an x and a (x-2). Those two are already covered before. So nothing more to bring down. So for LCD you are left with x * (x-2)

OpenStudy (ranga):

Let us try an example with numbers. Find the LCD of 1/3, 1/4 and 1/12 1sr row : 3: 3 2nd row: 4: 4 3rd row: 12: 3 x 4 From the first row drop down 3. From second row drop down 4. The third row has 3 and 4. But we have already covered 3 and 4 before and so we don't have to do it again. So the LCD will be 3 x 4 = 12

OpenStudy (anonymous):

Ohhh ! I seee ! So once you bring the down number once before ... you don't need to bring it down again!

OpenStudy (ranga):

Exactly. The whole idea in finding LCD or LCM is to come up with the minimum number of factors that covers ALL rows.

OpenStudy (ranga):

In the number example above, the LCD is 3 x 4. The first row has a 3 and we have that covered in our LCD. The second row has a 4 and we have that covered in the LCD. The third row has 3 x 4. We have both those factors covered in our LCD. So 3 x 4 is our LCD. The least number of factors that includes all factors in each row.

OpenStudy (anonymous):

How would you solve: \[\left| 6x ^{2} \right| 2 \times 3 \times x^{2}\] \[\left| 2x \right| 2 \times 2 \times x\] \[\left| 3x \right| 3 \times x \]

OpenStudy (ranga):

What do you mean solve. It is not an equation.

OpenStudy (anonymous):

Sorry. 2 (3) = 6x^2

OpenStudy (ranga):

You can multiply 2 and 3 on the left. Then divide both sides by 6. What do you get?

OpenStudy (anonymous):

you get 1

OpenStudy (ranga):

1 on the left but you have to divide on the right too.

OpenStudy (anonymous):

x^2

OpenStudy (ranga):

1 = x^2 or x^2 = 1 Take square root on both sides and you get x = sqrt(1) = +1 or -1 So x = plus or minus 1 is the solution.

OpenStudy (anonymous):

ooooh okay ! Cool Cool ! Thanks!!!

OpenStudy (ranga):

You are welcome.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!