hard PS multi choice

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by lunarpower » Fri Sep 19, 2008 6:10 pm
cramya wrote:Ron,
Thanks!

Are we going with none i.e choice A) because I) and II) toegther is not listed as a choice.

abc-b^3=0;
b(ab-b^2) = 0(I can understand the dividing by b part we cant do since b cane be 0 but what about factoring it)

b=0 or ac = b^2

Is this wrong?? Please help me understand
no, that's exactly right.
but don't forget what the question prompt is asking: which of the following statements MUST be true?
since there are two factors, EITHER of the two solutions may be true - but neither one of them MUST be.
in other words, you know that either b = 0 or ac = b^2, as you wrote above. but the key word here is OR: if i give you either one of those statements individually, you don't know whether it's true.

always read the question prompt carefully!

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note in general, though: factoring is a very good opener in general on a problem like this. if you have a problem on which you're tempted to divide out a variable on both sides, just FACTOR IT OUT instead. that will save you from losing certain solutions.
Ron has been teaching various standardized tests for 20 years.

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Hard ? choice PS Multi Choice---

by J.E » Wed Oct 22, 2008 12:07 am
If ABC= B^3 THEN A=1, B=1, C=1 Therefor AC= B^2, ABC=B^3
Then, D is the answer. :D

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by stop@800 » Wed Oct 22, 2008 10:50 pm
cramya wrote:Ron,
Thanks!

Are we going with none i.e choice A) because I) and II) toegther is not listed as a choice.

abc-b^3=0;
b(ab-b^2) = 0(I can understand the dividing by b part we cant do since b cane be 0 but what about factoring it)

b=0 or ac = b^2

Is this wrong?? Please help me understand
The question asks "Must be true" so we can not select b=0 or ac = b^2

with could be true, we can select these two choices.

Hope this helps