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varun7nurav
- Senior | Next Rank: 100 Posts
- Posts: 33
- Joined: Fri Nov 19, 2010 8:07 am
Hi guys,
Need help with a small doubt. I have seen conflicting solutions on Kaplan.
Problem: (x^2 - 1)/(x+1) = 1
In such situations, it looks evident to cancel out the common term after we expand the numerator which gives us the result (x-1)=1 resulting in x = 2.
Agree!
However, I have another problem that shows that such canceling isn't a good idea.
(a^2- b^2) = (a-b)^2
Solution 1: By expanding both sides, we get (a+b)(a-b)=(a-b)(a-b), canceling out common terms i.e. (a-b), we get (a+b)=(a-b), further solving gives us b=0 (Unique value)
Solution 2: By expanding using formula on RHS gives us
a^2 - b^2= (a^2 -2ab + b^2)
Canceling out a^2, we get b= 0 or b= a (Not unique)
This is something I really am not sure about.
Plz suggest me whr am I missing out? Am i forgetting any rule here?? If so, how do I decide when to cancel and when not to cancel out like terms?
Awaiting your response.
Thank you!
Need help with a small doubt. I have seen conflicting solutions on Kaplan.
Problem: (x^2 - 1)/(x+1) = 1
In such situations, it looks evident to cancel out the common term after we expand the numerator which gives us the result (x-1)=1 resulting in x = 2.
Agree!
However, I have another problem that shows that such canceling isn't a good idea.
(a^2- b^2) = (a-b)^2
Solution 1: By expanding both sides, we get (a+b)(a-b)=(a-b)(a-b), canceling out common terms i.e. (a-b), we get (a+b)=(a-b), further solving gives us b=0 (Unique value)
Solution 2: By expanding using formula on RHS gives us
a^2 - b^2= (a^2 -2ab + b^2)
Canceling out a^2, we get b= 0 or b= a (Not unique)
This is something I really am not sure about.
Plz suggest me whr am I missing out? Am i forgetting any rule here?? If so, how do I decide when to cancel and when not to cancel out like terms?
Awaiting your response.
Thank you!
