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Is root{(x-3)^2}=3-x? [#permalink]
 02 Apr 2010, 13:45
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Is \sqrt{(x-3)^2}=3-x? (1) x\neq{3}(2) -x|x| >0
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Is \sqrt{(x-3)^2}=3-x? Remember: \sqrt{x^2}=|x|. Why? Couple of things: The point here is that square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}. So \sqrt{x^2}\geq{0}. But what does \sqrt{x^2} equal to? Let's consider following examples: If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive; If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive. So we got that: \sqrt{x^2}=x, if x\geq{0}; \sqrt{x^2}=-x, if x<0. What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x|Back to the original question:So \sqrt{(x-3)^2}=|x-3| and the question becomes is: |x-3|=3-x? When x>3, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true. When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Basically question asks is x\leq{3}? (1) x\neq{3}. Clearly insufficient. (2) -x|x| >0, basically this inequality implies that x<0, hence x<3. Sufficient. Answer: B. Hope it helps.
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I think there's a slight flaw in the solution. Is sqrt((x-3)^2) = |x-3|? Arent we eliminating the negative square root here? I think the answer is D. Substitute -5 as an example, LHS is + -8 and RHS is +8 (or am I missing something here?)
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iambroke wrote: I think there's a slight flaw in the solution. Is sqrt((x-3)^2) = |x-3|? Arent we eliminating the negative square root here?
I think the answer is D. Substitute -5 as an example, LHS is + -8 and RHS is +8
(or am I missing something here?) First of all \sqrt{x^2} does equal to |x|, so \sqrt{(x-3)^2}=|x-3|. Next GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root. That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4. And finally (1) can not be sufficient as if you try x=5\neq{3}, then LHS=\sqrt{(x-3)^2}=\sqrt{(5-3)^2}=2\neq{RHS}=3-x=3-5=-2. Hope it's clear.
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Quote: When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root. Ok, that was news to me, but loks like you are right, just read through the OG for the first time just to confirm this. I hope the OG covers all such limitations that the test itself imposes(If not, please let me know). Thanks for the clarification though.
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Re: Root (x - 3) squared [#permalink]
10 Aug 2010, 00:31
You have to remember that taking the root of a square effectively makes the term positive.
So if you choose numbers x =/= 3, f.e. x=4, the LHS = +1, but the RHS = -1
if you choose x = -4 then the LHS = +7 and the RHS also = +7.
NOT SUFFICIENT
Answer 1 would have to be x <= 3 to be sufficient alone because then the RHS remains zero or positive.
Last edited by freedom on 10 Aug 2010, 08:56, edited 3 times in total.
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Re: explaination to a DS question.....plz help [#permalink]
07 Dec 2010, 01:45
dc123 wrote: is ((X-3)^2)^1/2 = 3-X ?
1) X does not = 3
2) -X|X| > 0
Help me plzz...Want to Master the GMAT HAHA! ((X-3)^2)^1/2 = X-3 is X>=3 OR ((X-3)^2)^1/2 = 3-X if X<=3 st 1) x!=3, not sufficient as we dont know if x is greater than 3 or less than 3 st 2) -X|X| > 0 .. as |X| is always postive, X has to be negative. so ((X-3)^2)^1/2 = 3-X. sufficient B
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Re: explaination to a DS question.....plz help [#permalink]
07 Dec 2010, 02:42
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dc123 wrote: is ((X-3)^2)^1/2 = 3-X ?
1) X does not = 3
2) -X|X| > 0
Help me plzz...Want to Master the GMAT HAHA! For {\sqrt{(X-3)^2} = 3-X to be true, 3-X will always have to be positive or 0, because in GMAT a root can not be negetive. So 3-X>0 or X<3. S1: X =! 3, but is it less than 3? we dont know. Not sufficient. S2: -X|X|>0, so either both ( -X and |X|) are -ive or both are +ive. |X| can not be negetive so -X must be +ive, or X must be negetive, which would clearly be less than 3. Sufficient.Answer: B
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Re: explaination to a DS question.....plz help [#permalink]
07 Dec 2010, 23:48
vaibhavtripathi wrote: dc123 wrote: is ((X-3)^2)^1/2 = 3-X ?
1) X does not = 3
2) -X|X| > 0
Help me plzz...Want to Master the GMAT HAHA! For {\sqrt{(X-3)^2} = 3-X to be true, 3-X will always have to be positive or 0, because in GMAT a root can not be negetive. So 3-X>0 or X<3. S1: X =! 3, but is it less than 3? we dont know. Not sufficient. S2: -X|X|>0, so either both ( -X and |X|) are -ive or both are +ive. |X| can not be negetive so -X must be +ive, or X must be negetive, which would clearly be less than 3. Sufficient.Answer: BI like your simple way of solving but how do you know ((X-3)^2)^1/2 = 3-X ? the question is asking us if this is true not to take it as true. Unless you believe that in order for the equation to be true this has to be the case.
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mrcrescentfresh wrote: I am having a hard time grasping why we cannot simplify the problem as:
((X-3)^2)^1/2 = (3 - X) to (X - 3)^2 = (3 - X)^2?
I know I have worked problems before where we have been able to solve it by squaring both sides, but when done in this scenario the answer is entirely different than the OA.
Please help. That is because if the question says: Is 5 = -5? And you do not know but you square both sides and get 25 = 25 Can you say then that 5 = -5? No! I understand that you would have successfully used the technique of squaring both sides before but that would be in conditions like these: Given equation: \sqrt{X} = 3Squaring both sides: X = 9 Here you already know that the equation holds so you can square it. It will still hold. This is like saying: It is given that 5 = 5. Squaring both sides, 25 = 25 which is true. This question is similar to the first case. It is asked whether \sqrt{((X-3)^2)} = (3 - X)? LHS is positive because \sqrt{((X-3)^2)} = |X-3|and by definition of mod, we know that |X| = X if X is positive or zero and -X if X is negative (or zero). Since |X-3| = - (X - 3), we can say that X - 3 <= 0 or that X <= 3 So the question is: Is X <= 3? Stmnt 1 not sufficient. But stmnt 2 says -X|X| > 0 This means -X|X| is positive. Since |X| will be positive, X must be negative to get rid of the extra negative sign in front. So this statement tells us that X < 0. Then X must be definitely less than 3. Sufficient.
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iambroke wrote: Quote: When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root. Ok, that was news to me, but loks like you are right, just read through the OG for the first time just to confirm this. I hope the OG covers all such limitations that the test itself imposes(If not, please let me know). Thanks for the clarification though. This is not GMAT specific! It is the convention. Given x = \sqrt{9}, x = 3 only Given x^2 = 9, x can be 3 or -3
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Re: explaination to a DS question.....plz help [#permalink]
09 Dec 2010, 05:38
vaibhavtripathi wrote: dc123 wrote: is ((X-3)^2)^1/2 = 3-X ?
1) X does not = 3
2) -X|X| > 0
Help me plzz...Want to Master the GMAT HAHA! For {\sqrt{(X-3)^2} = 3-X to be true, 3-X will always have to be positive or 0, because in GMAT a root can not be negetive. So 3-X>0 or X<3. S1: X =! 3, but is it less than 3? we dont know. Not sufficient. S2: -X|X|>0, so either both ( -X and |X|) are -ive or both are +ive. |X| can not be negetive so -X must be +ive, or X must be negetive, which would clearly be less than 3. Sufficient.Answer: BA word of caution: This works only because left side of the equation has X - 3. If I twist the question a little: Is {\sqrt{(X-1)^2} = 3-XI get: |X - 1| = 3 - X This has only one solution X = 2. So now the question is: Is X = 2? (Not Is X < 3?) It is a more specific question now.
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Bunuel wrote:
Remember: \sqrt{x^2}=|x|. Why?
I am confused. I thought there was some kind of agreement that square root sign indicates a positive number. (For example, \sqrt{25}=5 and not "-5"). On the other hand, if x^2=25, then x can be 5 or -5.
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Fijisurf wrote: Bunuel wrote:
Remember: \sqrt{x^2}=|x|. Why?
I am confused. I thought there was some kind of agreement that square root sign indicates a positive number. (For example, \sqrt{25}=5 and not "-5"). On the other hand, if x^2=25, then x can be 5 or -5. Frankly speaking I don't see contradiction between the fact that \sqrt{x^2}=|x| and that even roots give non-negative result, as absolute value also gives non-negative result. Anyway: Square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}, or \sqrt{x}\geq{0}. So, when the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the non-negative root. That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.On the other hand absolute value is also non-negative |some \ expression|\geq{0}, or |x|\geq{0} (absolute value of x, |x|, is a distance between point x on a number line and zero, so distance can not be negative). As for \sqrt{x^2}=|x|: Consider following examples: If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive; If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive. So we got that: \sqrt{x^2}=x, if x\geq{0}; \sqrt{x^2}=-x, if x<0. Which exactly what the absolute value function does: |x|=x if x\geq{0} and |x|=-x if x\leq{0}. So, \sqrt{x^2}=|x|. Is this still confusing?
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Re: Algebra DS question [#permalink]
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rahuljaiswal wrote: 20. Is \sqrt{(x-3)^2} = 3 - x ? (1) x ≠ 3 (2) – x | x | > 0 Did not understand Statement 2 at all..is that a product? Kindly help P.S: the question reads as " Is square root of square of (x-3) equal to (3 - x) ?" Hi! Statement (2) reads as "the product of (negative x) and (absolute value of x) is positive". Since |x| is always non-negative (i.e. |x|=0 when x=0 and is positive for all other values of x), in order for statement (2) to be true, x must be negative, giving us: -(-) * |-| > 0 + * + > 0 which is true. So, basically, (2) is a fancy way of saying: x < 0. Now that you know how to interpret (2), try the question again - if you still need a hand, let us know where you get stuck!
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GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ?? [#permalink]
15 Jan 2013, 23:01
There is a DS question on GMAT Prep1 test that I came across: \sqrt{(x-3)^2} = 3-x1. x !=3 2. -x|x| >0. I looked at Bunuel's solution: is-sqrt-x-3-2-3-x-1-x-not-equal-to-3-2-x-x-62992.htmlHe says that sqrt(expression) will always return >=0, is this right? I thought that sqrt(25) for instance would be +- 5, and not just 5. Can anyone please help me with understanding this? I do not agree with the explanation, please help.
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Re: GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ?? [#permalink]
15 Jan 2013, 23:22
soods26 wrote: There is a DS question on GMAT Prep1 test that I came across: \sqrt{(x-3)^2} = 3-x1. x !=3 2. -x|x| >0. I looked at Bunuel's solution: is-sqrt-x-3-2-3-x-1-x-not-equal-to-3-2-x-x-62992.htmlHe says that sqrt(expression) will always return >=0, is this right? I thought that sqrt(25) for instance would be +- 5, and not just 5. Can anyone please help me with understanding this? I do not agree with the explanation, please help. As far as gmatclub is concerned, when Bunuel says something, I take it as my final source for that bit of information. So you can rest assured that he is right. Coming to your question if x^2 = 25, then x = + or - 5 However, if x = \sqrt{25}, then x = 5
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Re: GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ?? [#permalink]
15 Jan 2013, 23:42
MacFauz wrote: As far as gmatclub is concerned, when Bunuel says something, I take it as my final source for that bit of information. So you can rest assured that he is right.
Coming to your question
if x^2 = 25, then x = + or - 5
However, if x = \sqrt{25}, then x = 5 Thank you for your reply, I have been around in hiding on GMAT Club long enough to know the "demi-god" that Bunuel is  However coming to: x^2 = 25, the next step in solving the above equation, is taking roots both sizes, which gives us: x = \sqrt{25}, which according to you above gives x = 5 and not +- 5. So there lies my confusion ...
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Re: GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ?? [#permalink]
15 Jan 2013, 23:51
soods26 wrote: MacFauz wrote: As far as gmatclub is concerned, when Bunuel says something, I take it as my final source for that bit of information. So you can rest assured that he is right.
Coming to your question
if x^2 = 25, then x = + or - 5
However, if x = \sqrt{25}, then x = 5 Thank you for your reply, I have been around in hiding on GMAT Club long enough to know the "demi-god" that Bunuel is However coming to: x^2 = 25, the next step in solving the above equation, is taking roots both sizes, which gives us: x = \sqrt{25}, which according to you above gives x = 5 and not +- 5. So there lies my confusion ... does not mean the same as x = \sqrt{25}The radical(square root) symbol is just another operator similar to a + or -. Hence \sqrt{25} can only take one value and that is the positive one. While x^2 = 25 is an equation that needs to be solved, x = \sqrt{25} emphatically gives the value of x and is just a roundabout way of saying x = 5. Hope it is clear.
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Re: GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ?? [#permalink]
16 Jan 2013, 03:34
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Re: GMAT Prep1 Question: IS sqrt(an expression) >= 0 always ??
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