imo D
stem: x-3 =3-x
i.e. x=3

stmt 1: x=/ 3, ans no suff
stmt2: x=-ve, not 3 hence suff

Well I finally got it. There is a similar question in OG

sq root of x-3 can be either x-3 or 3-x. the only way you can eliminate one of the two choice is if u know

1) if value of x is +ve or -ve or
2) if 3 is positive or negative

Now look at the statement

1) x not equal 3. So what? X can still be positive or negative.

2) The only way this statement is true is if x is negative. Since the absolute value of x is always positive. Ans -x lxl > 0.

ri2007 wrote:Well I finally got it. There is a similar question in OG

sq root of x-3 can be either x-3 or 3-x. the only way you can eliminate one of the two choice is if u know

1) if value of x is +ve or -ve or
2) if 3 is positive or negative

Now look at the statement

1) x not equal 3. So what? X can still be positive or negative.

2) The only way this statement is true is if x is negative. Since the absolute value of x is always positive. Ans -x lxl > 0.

Not quite there in your answer.

The original question stem equation sqrt((x-3)^2)=3-x can be rewritten as |x-3|=3-x

-note that if you are squaring a negative number and then taking the squareroot of that number, you get a positive number, hence AbsoluteValue(x-3) or |x-3|.

1) plugging positive and negative values will get you different values, therefore insufficient

2) -x|x|>0 implies that x<0>0 which is -1>0 which does not follow the inequality. Plug -1 for x and you will get that x<0. This limits what we can plug into the original equation above. Now plug in any value less than zero and the above equation is true. Therefore, Sufficient (B)

2) -x|x|>0 implies that x<0>0 which is -1>0 which does not follow the inequality. Plug -1 for x and you will get that x<0. This limits what we can plug into the original equation above. Now plug in any value less than zero and the above equation is true. Therefore, Sufficient (B)[/quote]

I did not get step(2) that you were explaining. Can you debrief on this further.

ri2007 whats the similar OG sum? Can you tell me the problem no?

gmatrant wrote:

2) -x|x|>0 implies that x<0>0 which is -1>0 which does not follow the inequality. Plug -1 for x and you will get that x<0. This limits what we can plug into the original equation above. Now plug in any value less than zero and the above equation is true. Therefore, Sufficient (B)

I did not get step(2) that you were explaining. Can you debrief on this further.

ri2007 whats the similar OG sum? Can you tell me the problem no?

sorry...that's a typo. it should say "x<0" (there is something wrong with the software on this forum I just tried to write the "x<0" and it keeps showing up as x<0>0. Just figured that one out. the "<" and ">" are parts of html tags so they screw with the forum software unless you disable html in your posts)

-x|x|>0

to fulfill this inequality all values of x must be negative. If we plug +1 for x we get -(+1)|1|>0, so we have -1(1)>0, then -1>0. This does not fulfill the inequality. This will occur for any positive vale of x. Subsequently, plugging zero gives 0>0, which also does not work. Plugging in any negative value, however, will work. -(-1)|-1|>0. We get 1(1)>0 or 1>0, which is true.

enlightenment wrote:

gmatrant wrote:

2) -x|x|>0 implies that x<0>0 which is -1>0 which does not follow the inequality. Plug -1 for x and you will get that x<0. This limits what we can plug into the original equation above. Now plug in any value less than zero and the above equation is true. Therefore, Sufficient (B)

I did not get step(2) that you were explaining. Can you debrief on this further.

ri2007 whats the similar OG sum? Can you tell me the problem no?

sorry...that's a typo. it should say "x<0" (there is something wrong with the software on this forum I just tried to write the "x<0" and it keeps showing up as x<0>0. Just figured that one out. the "<and>0

to fulfill this inequality all values of x must be negative. If we plug +1 for x we get -(+1)|1|>0, so we have -1(1)>0, then -1>0. This does not fulfill the inequality. This will occur for any positive vale of x. Subsequently, plugging zero gives 0>0, which also does not work. Plugging in any negative value, however, will work. -(-1)|-1|>0. We get 1(1)>0 or 1>0, which is true.

Thanks for you post..enlightenment.. but I am still stuck with respect to how B will suffice.

As per your logic x has to be negative since -x|x| >0. Thats my understanding too. After after this i fail to understand as to how do you come to a conclusion that since x is negative LHS will be equal to RHS.

LHS is sqrt([x-3]^2)
So lets take any negative value for x, lets say x is -4
so sqrt([-4-3]^2) = sqrt(49) = + or - 7

RHS is 3-x , x is -4
so RHS is 7.
+ or -7 is not equal to 7.
hence E.

This is where I fail to understand how B holds good. Please comment.
Need some enlightenment...

I am seeing quite a few problems like this but I am not able to solve this.
Can anyone clarify my doubt with respect to my previous problem related to the problem

@gmatrant ... you say in the previous post that sqrt(49) is + or - 7 .. this is not true .. Remember the sqrt of a number is always positive, so the sqrt of 49 is 7, that is why it is said that sqrt((x-3)^2) is equivalent to mod(x-3),

7 and -7 are the roots of the equation x^2 - 49 =0 .. hope this clears up your doubts ..

Regards