For all integers n, n*=n(n-1). What is the value of x*? I)

Question

For all integers n, n*=n(n-1). What is the value of x*?

I) x*=x
II) (x-1)*=x-2

MA · Answer

go with B. Explain later.........

FN · Answer

B it is

1) x*=x

plug numbers

lets pick 0 and 2

0(0-1)=0 satisfy the stem

2(2-1)=2 satisfy the stem

1 is insuff


2)
plug numbers and you will see that 2 is the only option

2 is sufficient!

kevinw · Answer

what happens is you plug 3 into II) ???

jma123 · Answer

I thought this could be done using algebra:

S1: x* = x
x(x-1) = x
x-1=1
x=2
sufficient

S2: 
(x-1)(x-2)=x-2
x-1=1
x=2
suff

what am I doing wrong?

HongHu · Answer

You thinking process is correct and very good, but you made an algebra mistake.

S1: x* = x
x(x-1) = x
x(x-1)-x=0
x(x-2)=0
x=0 or x=2

non unique solution
insufficient

S2: 
(x-1)(x-2)=x-2
(x-1)(x-2)-(x-2)=0
(x-2)(x-2)=0
x=2
unique solution
sufficient

jma123 · Answer

Hmm, I had thought I could divide through:

S1: x(x-1) = x
 divide both side by x --> x-1 = 1

S2: (x-1)(x-2) = (x-2)
 divide both sides by x-2 --> x-1 = 1

So to be on the safe side for the test, that we should multiply out instead of cancelling out?

Arsene_Wenger · Answer

I solved this problem and arrived at (D) in about 30 seconds, even though i agree it is (B). I was glad because to me, i may have made up a lot of time by answering it in 30 secs. But what happens when one faces these questions on matchday? Wont it seem to be a waste of time plugging in nos. (-2, -1, 0, 1, 2, 1/4, etc) when simple algebra seems to work in solving the problem?

banerjeea_98 · Answer

my personal opinion, if u see eqn always use algebra, my personal opinion is plugging no is waste of time, more importantly u always have doubts whether u covered all possinbilities, algebra sol is bullet proof. Offcourse, there r cases when only plugging will work.

MA · Answer

Absoutly right. I agree with Banerjee. Plugging in is last resort.

Arsene_Wenger · Answer

This was also my approach. Can anyone tell why we cannot divide both sides by a common term as in jma's approach in S1?

vprabhala · Answer

I too did the same as jma123. can somebody explain why i shouldn't be doing that/

HongHu · Answer

You need to be very careful when you do algebra derivations. One of the common mistake is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.

Example:

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.

Arsene_Wenger · Answer

Thanks HongHu

For all integers n, n=n(n-1). What is the value of x? I)

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