Isnt the answer E
case 1: dosent help
case 2:
we know from this that x has to be negative, so lets take x as -3
LHS sqrt[(-3-3)^2]
it can be or + or - 6
RHS 3- (-3) = 6
So even if we combine the two its not possible to say what the answer is?
Where am i going wrong.
Tricky square root problem
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 36
- Joined: Sat Jul 21, 2007 10:55 am
- Location: India
- Thanked: 1 times
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
stem: x-3 =3-x
i.e. x=3
stmt 1: x=/ 3, ans no suff
stmt2: x=-ve, not 3 hence suff
Hi friends ! Just started GMAT prep. Hope to share thoughts, experience, joy and agony of GMAT prep journey.
Regards; neeraj
Regards; neeraj
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.
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.
-
- Junior | Next Rank: 30 Posts
- Posts: 20
- Joined: Sat Nov 10, 2007 10:34 pm
Not quite there in your answer.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.
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)
-
- Master | Next Rank: 500 Posts
- Posts: 416
- Joined: Wed Oct 03, 2007 9:08 am
- Thanked: 10 times
- Followed by:1 members
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?
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?
-
- Junior | Next Rank: 30 Posts
- Posts: 20
- Joined: Sat Nov 10, 2007 10:34 pm
gmatrant wrote:I did not get step(2) that you were explaining. Can you debrief on this further.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)
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.
-
- Master | Next Rank: 500 Posts
- Posts: 416
- Joined: Wed Oct 03, 2007 9:08 am
- Thanked: 10 times
- Followed by:1 members
Thanks for you post..enlightenment.. but I am still stuck with respect to how B will suffice.enlightenment wrote:gmatrant wrote:I did not get step(2) that you were explaining. Can you debrief on this further.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)
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.
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...
- gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
@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
7 and -7 are the roots of the equation x^2 - 49 =0 .. hope this clears up your doubts ..
Regards