If x(x - 5)(x + 2) = 0, is x negative?
(1) x2 – 7x ≠ 0
(2) x2 –2x –15 ≠ 0

First of all, you see that x can be 0 or 5 or -2.

1) x^2-7x≠0 is also x(x-7)≠0
So x≠0 and x≠7
For your equation x(x - 5)(x + 2) = 0, x can still be 5 or -2 but not 0. The problem is that 5 is positive and -2 negative so you cannot answer the question. It is insufficient.

2) x^2 –2x –15 ≠ 0 is also (x-5)(x+3)≠0
So x≠5 and x≠-3
For your equation x(x - 5)(x + 2) = 0, x can still be 0 or -2 but not 5.
The problem is that 0 is neither negative or positive, so we cannot tell if x is positive or negative, we do not know.If 0 were negative, we could have told it is sufficient but it is not the case so we can't. It is insufficient.

1) and 2)
x≠5
x≠-3
x≠0
x≠7
The only solution is x=-2

The answer is C.
Tell me if wrong or problems.

pepeprepa wrote:If x(x - 5)(x + 2) = 0, is x negative?
(1) x2 – 7x ≠ 0
(2) x2 –2x –15 ≠ 0

First of all, you see that x can be 0 or 5 or -2.

1) x^2-7x≠0 is also x(x-7)≠0
So x≠0 and x≠7
For your equation x(x - 5)(x + 2) = 0, x can still be 5 or -2 but not 0. The problem is that 5 is positive and -2 negative so you cannot answer the question. It is insufficient.

2) x^2 –2x –15 ≠ 0 is also (x-5)(x+3)≠0
So x≠5 and x≠-3
For your equation x(x - 5)(x + 2) = 0, x can still be 0 or -2 but not 5.
The problem is that 0 is neither negative or positive, so we cannot tell if x is positive or negative, we do not know.If 0 were negative, we could have told it is sufficient but it is not the case so we can't. It is insufficient.

1) and 2)
x≠5
x≠-3
x≠0
x≠7
The only solution is x=-2

The answer is C.
Tell me if wrong or problems.

I think answer is 'B'.
I agree with you on stmt 1..but on stm2, I tried substituting values of x in stmt2 instead of solving it..
From question, x=5 or -3
Susbtitute x=5 in stmt2...25-10-15 = 0 ..hence x cannot be 5
now substitute x=-2 in stmt2, 4+4-15=7 <> 0..hence x =-2..negative..sufficient...

what is the OA?

bha wrote:

pepeprepa wrote:If x(x - 5)(x + 2) = 0, is x negative?
(1) x2 – 7x &#8800; 0
(2) x2 –2x –15 &#8800; 0

First of all, you see that x can be 0 or 5 or -2.

1) x^2-7x&#8800;0 is also x(x-7)&#8800;0
So x&#8800;0 and x&#8800;7
For your equation x(x - 5)(x + 2) = 0, x can still be 5 or -2 but not 0. The problem is that 5 is positive and -2 negative so you cannot answer the question. It is insufficient.

2) x^2 –2x –15 &#8800; 0 is also (x-5)(x+3)&#8800;0
So x&#8800;5 and x&#8800;-3
For your equation x(x - 5)(x + 2) = 0, x can still be 0 or -2 but not 5.
The problem is that 0 is neither negative or positive, so we cannot tell if x is positive or negative, we do not know.If 0 were negative, we could have told it is sufficient but it is not the case so we can't. It is insufficient.

1) and 2)
x&#8800;5
x&#8800;-3
x&#8800;0
x&#8800;7
The only solution is x=-2

The answer is C.
Tell me if wrong or problems.

I think answer is 'B'.
I agree with you on stmt 1..but on stm2, I tried substituting values of x in stmt2 instead of solving it..
From question, x=5 or -3
Susbtitute x=5 in stmt2...25-10-15 = 0 ..hence x cannot be 5
now substitute x=-2 in stmt2, 4+4-15=7 <> 0..hence x =-2..negative..sufficient...

what is the OA?

oops..my bad..I read the question as (x-5)(x+2)=0..didn't see the x at the beginning... :mrgreen:

Yes OA is a C. Tks
What a simple solution.