For all integers n, n* = n(n – 1). what is the value of x* when x is an integer?

(1) x* = x --> \(x(x-1)=x\) --> \(x(x-1)-x=0\) --> \(x(x-2)=0\) --> \(x=0\) or \(x=2\). Thus \(x*=0(0-1)=0\) or \(x*=2(2-1)=2\). Not sufficient.

If you divide (reduce) \(x(x-1)=x\) by x you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution (notice that both x=0 AND x=2 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.


(2) (x – 1)* = x – 2 --> \((x-1)(x-2)=x-2\) --> \((x-2)^2=0\) --> \(x=2\). Sufficient.

Answer: B.

Hope it's clear.
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HI Bunuel,

How are you getting (x-2)^2 = 0 in second statemnt?

Thanks.

\((x-1)(x-2)=x-2\);

\((x-1)(x-2)-(x-2)=0\);

\((x-2)(x-1-1)=0\);

\((x-2)(x-2)=0\).

Hope it's clear.
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Please tag this to DS instead of PS. Thank you
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Done. Thanks for the information.
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My questions is whether you are algebraically allowed to divide by X if you have the equation: x(x-1) = x. Dividing by x yields x-1 = 1 whereas not dividing yields x^2 - 2x.
you need to take into consideration that x might be 0!!! 0 is an integer too!

my approach:
1. x* = x
x(x-1) = x
x^2 -2x = 0
x^2 = 2x
x can be 2 or x can be 0 - 2 outcomes; not sufficient. AD are out.

2. (x-1)* = x-2
(x-1)(x-2) = x-2
x^2 -2x-x+2 = x-2
x^2 - 4x +4 = 0
(x-2)^2 = 0
only 1 solution: x-2 = 0, and x=2.
sufficient. answer is B.

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