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Re: M03-19 [#permalink]
Bunuel wrote:
Official Solution:


In order \(abc = 0\) to be true at least one of the unknowns must be zero.

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient.

(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)

\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\);

\(b=c-c\);

\(b=0\). Sufficient.


Answer: B


I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=-b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.).

so it should be E.
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Re: M03-19 [#permalink]
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suhasreddy wrote:
Bunuel wrote:
Official Solution:


In order \(abc = 0\) to be true at least one of the unknowns must be zero.

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient.

(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)

\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\);

\(b=c-c\);

\(b=0\). Sufficient.


Answer: B


I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=-b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.).

so it should be E.


That's not true. If a+b were 0, then \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) would not be correct for any real a, b, and c.

When having something like x^2/x=y you can ALWAYS reduce by x and write x = y.

The answer is B.
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Re: M03-19 [#permalink]
I guess what I'm saying is that I've never seen this operation before. I believe it, because I basically get the same answer the long way:

(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\))
...and then the foil method

If it's a valid shortcut then I wonder if there is a proof or something for it so I know when I can use it in other situations.
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Re: M03-19 [#permalink]
Bunuel wrote:
Official Solution:


In order \(abc = 0\) to be true at least one of the unknowns must be zero.

(1) \(a^2 = 2a\). Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient.

(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)

\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\);

\(b=c-c\);

\(b=0\). Sufficient.


Answer: B


I'm getting confused on a particular step, when \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) is being simplified to \(b= \frac{c*(a+b)^2}{(a+b)^2} - c\).
When solving it I'm getting \(b= \frac{c*(a^2+b^2)}{(a+b)^2} - c\). And I'm not sure how \((a^2+b^2)\) is being written as \((a+b)^2\), as \((a+b)^2 = {a^2+2ab+b^2}\)
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Re: M03-19 [#permalink]
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banerjr1 wrote:

I'm getting confused on a particular step, when \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\) is being simplified to \(b= \frac{c*(a+b)^2}{(a+b)^2} - c\).
When solving it I'm getting \(b= \frac{c*(a^2+b^2)}{(a+b)^2} - c\). And I'm not sure how \((a^2+b^2)\) is being written as \((a+b)^2\), as \((a+b)^2 = {a^2+2ab+b^2}\)


Let's do this with some intermediate steps so the math is more clear:

\(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)

Factor out \(\sqrt{c}\) from \(a*\sqrt{c}+b*\sqrt{c}\)

\(b= \frac{(\sqrt{c}*(a+b))^2}{a^2+2ab+b^2} - c\)

Distribute the square to \(\sqrt{c}\) and \(a+b\) (This is where you made your mistake — we can distribute a square to terms that are being multiplied together, but not terms that are being added together. This is why we get \((a+b)^2\) and not \(a^2 + b^2\).)

\(b= \frac{\sqrt{c}^2*(a+b)^2}{a^2+2ab+b^2} - c\)

\(b= \frac{c*(a+b)^2}{a^2+2ab+b^2} - c\)

Either expand \((a+b)^2\) in numerator to \(a^2+2ab+b^2\) OR factor \(a^2+2ab+b^2\) in denominator to \((a+b)^2\)

\(b= \frac{c*(a^2+2ab+b^2)}{a^2+2ab+b^2} - c\) OR \(b= \frac{c*(a+b)^2}{(a+b)^2} - c\)

Cancel like terms from numerator and denominator (either \(a^2+2ab+b^2\) or \((a+b)^2\))

\(b= c - c\)

\(b= 0\)

So you're totally correct that \(a^2 + b^2\) doesn't equal \((a+b)^2\), but we should never get \(a^2 + b^2\) based on exponent rules for distributing exponents.
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Re: M03-19 [#permalink]
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Bunuel wrote:
Is \(abc = 0\)?


(1) \(a^2 = 2a\)

(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)


(1) \(a^2 = 2a\)\(=a^2-2a=0\)

\(= a(a-2)=0\)

\(or, \ a=0, a=2\)

\(If \ a=0 \ then \ abc=0 \ and \ a=2 \ then \ abc \ not \ equal \ to \ 0 \ Insufficient. \)

(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)

\(b= \frac{\sqrt{c}^2(a+b)^2}{(a+b)^2}-c\)

\(b=\frac{c(a+b)^2}{(a+b)^2}-c\)

\(b=c-c=0\)

Sufficient.

The answer is B
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Re: M03-19 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M03-19 [#permalink]
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Similar PS question to practice:

https://gmatclub.com/forum/m41-429461.html
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