itsagulati wrote:
For all integers n, n* = n(n – 1). what is the value of x* when x is an integer?
(1) x* = x
(2) (x – 1)* = x – 2
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.
Changes the answer...
Thanks!
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.