iliavko
Bunuel,
A question here, why can't we assume that since a+b=8 then a^2+b^2=64? same as a^2-b^2=16, then sq root both sides and get a-b=4
Thank you!
Let me try to answer.
If you are given \(a+b=8\), then you CAN NOT write it as \(a^2+b^2=64\). The reason is that you must square both sides COMPLETELY,
i.e.
\(a+b=8\) ---> squaring both sides ---> \((a+b)^2=8^2\) ---> \(a^2+2ab+b^2=64\)
Think of it this way, if I tell you that a+b=8 ----> I can have 1 case for (a,b) as a=7, b=1 , does \(a^2+b^2=64\)? NO.
Also, I dont know where you saw that then \(a^2-b^2=16\) --->\(a-b=4\) but this is again incorrect.
If you have \(a^2-b^2=16\) ---> the only correct manipulation for this expression is \((a+b)(a-b)=16\) and nothing else.
Again, if \(a^2-b^2=16\), I can have \(a= \sqrt{17}\) and \(b=1\), now does a-b = 4 ? NO
The formulae that I have used above are (learn these as they are one of the most used formulae in GMAT Quant):
\((a + b) = a^2 + 2ab + b^2\)
\((a - b) = a^2 - 2ab + b^2\)
\(a^2-b^2 = (a+b)*(a-b)\)
FYI,
1. If you are given a+b = 8 , then a^2+b^2=64 ONLY IF 2ab = 0 ---> ab=0 ---> either a or b=0 or even BOTH a,b =0. Thus the only case when if a+b = 8 , then a^2+b^2=64, is when either a OR b =0.
2. If you are given a^2-b^2 = 16 ---> a-b=4 , ONLY POSSIBLE when b=0 and a=4Hope this helps.