Kaplan 800 question Practice set 6, question #13

slimsohn wrote:Can someone help me figure out where I'm going wrong here:

For all nonzero integers n, n*=(n+2)/n . What is the value of x?

(1) x* = x
(2) x* = -2-x


I know its not (1), but I think it could be (2) because

x* = (x+2)/x = -2-x = -1 (2+x)
if you cancel (x+2) from both sides

1/x = -1 --> x = -1.

But if you expand it without cancelling, you end up with x^2 + 3x + 2, which I know to has x=-1, -2 as soln...

I guess, my question is, why cant I just cancel the x+2 after I factor a -1 out?

thanks.

any eqn of the form x^n + bx^(n-1)+cx^(n-2)+..... = 0 will have n roots ( roots are the value of x at which the eqn has a value of 0 )... so when we are dealing with a quadratic eqn (that is n=2), as it is in this case, we have to remember that there will always be 2 values for x at which the eqn has a value of 0....

So u can cancel out x+2 if u want to.. but always remember that the value of x u get after doing this is just one of the value of x that satisfies the eqn...

Gmat has used this property of a quadratic eqn cleverly in DS questions..
So whenevr u come across a question such as x(x+2)=x or (x+2)/x=(x+2).... remember that x will have 2 values that satisfy the eqn .. write it down if necessary...... so for the above ds the answer shuld be c ... hope that helps