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# M03-19

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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.

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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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.

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.

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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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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.

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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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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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.

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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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Similar PS question to practice:

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