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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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22 Nov 2014, 07:20
hanschris5 wrote: Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps. can u pls help in decoding option B where , you say x <0 i mean xx> 0 , could you elaborate on this . thanks. Sure. We have that \(xx >0\), so the product of two multiples x and x is positive. Now, we know that x must be positive, hence x must also be positive for the product to be positive, thus x > 0, which is the same as x < 0. Hope it's clear.
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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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22 Nov 2014, 08:26
okay gr8 ... I have a doubt about this we have sqrt of x is always greater than equal to 0 . Now same concept is what we applied on the above problem. But there is one problem that u explained on a GMAT PREP question . Let's consider following examples: I am concern about the second one when x=5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=x I tried posting this on the BUT was not accepting the expression ... havent yet learnt to use expression ... thanks in advance. Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps.



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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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22 Nov 2014, 09:09
hanschris5 wrote: okay gr8 ... I have a doubt about this we have sqrt of x is always greater than equal to 0 . Now same concept is what we applied on the above problem. But there is one problem that u explained on a GMAT PREP question . Let's consider following examples: I am concern about the second one when x=5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=x I tried posting this on the BUT was not accepting the expression ... havent yet learnt to use expression ... thanks in advance. Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps. \(\sqrt{x^2}=\sqrt{25}=5\) > x = 5 or x = 5. The square root gives positive result but x itself can be 5 or 5. What is confusing there?
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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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22 Nov 2014, 10:26
okay so its only do with the result ie the x that comes out as result after the operation ... whatever it may have been inside the expression...
thanks. doubt cleared.



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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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27 Nov 2015, 09:11
Given to us \(\sqrt{(x3)^2} = 3x\)
As a general principle, whenever we are given square root of a square of a number, we can have 2 solutions for such a number. It is illustrated below:
\(\sqrt{x^2} = x (if x>=0)\) \(\sqrt{x^2} = x (if x<=0)\) This square root of a square of a number is essentially the modulus of number and thus always gives a positive answer.
So in questions like the one illustrating the above principle we have to consider both the possibilities.
Now, for the given equation:
The two possibilities are
1st possibility : \(\sqrt{(x3)^2} = x3\) (if x>=3) 2nd possibility : \(\sqrt{(x3)^2} = 3x\) (if x<=3)
Now considering the given statements:
Using Statement 1
if \(x\neq{3}\)
This statement satisfies both the possibilities Hence we cannot reach to a single possibility, so statement 1 is not sufficient.
Using Statement 2
\(xx > 0\)
For this condition to be true, x needs to be necessarily less than 0.
Now if x<0, then it definitely is less than 3, hence will satisy the 2nd possibility, thus, giving a distcict answer.
Hence statement 2 is sufficent to answer \(\sqrt{(x3)^2} = 3x\)
So the answer is B.



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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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26 Jan 2017, 19:55
Be definition: √(x²) = x. xy is the DISTANCE between x and y. The DISTANCE between two numbers must be greater than or equal to 0. The question stem above, rephrased: Is x3 = 3x? In words: Is the DISTANCE between x and 3 equal to the DIFFERENCE of 3 and x? The answer will be YES if the DIFFERENCE of 3 and x is greater than or equal to 0: 3x≥0 x≤3. The question stem rephrased: Is x≤3? Statement 1: x is not equal to 3. It is possible that x<3 or that x>3. INSUFFICIENT. Statement 2: x*x > 0 . Thus, the lefthand side must be positive*positive or negative*negative. Since x cannot be negative, both factors on the lefthand side must be positive. Thus: x>0 x<0. Since x<0, we know that x≤3. SUFFICIENT. The correct answer is B.
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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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02 Feb 2017, 02:51
hanschris5 wrote: Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps. can u pls help in decoding option B where , you say x <0 i mean  xx> 0 , could you elaborate on this . thanks. Whatever you put into the x, it always gives you positive value. If you multiply 2 positive or 2 negative values, it always gives you positive result, which is ''some expression''>0. Here, x already gives you positive value. So, the green part has to be negative to legit the statement. Here, is test case: x*positive value>0 > (10)*positive value>0 >positive value*positive value>0 So, to legit the statement 2, x must be negative. The statement is sufficient.
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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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23 Apr 2017, 20:01
Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps. Hi Bunuel, From \(x3=3x\), since we know x3 is positive, can we not go one step further and simplify the equation to: x3=3x =>x=3? What am I doing wrong?



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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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23 Apr 2017, 22:44
puto wrote: Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps. Hi Bunuel, From \(x3=3x\), since we know x3 is positive, can we not go one step further and simplify the equation to: x3=3x =>x=3? What am I doing wrong? I tried to explain exactly this point in may solutions. From x  3 = 3  x, when x  3 <= 0. If it's still unclear you should go through theory about absolute values: Theory on Absolute Values: http://gmatclub.com/forum/mathabsolute ... 86462.htmlAbsolute value tips: http://gmatclub.com/forum/absolutevalu ... 75002.htmlHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Is sqrt ((x3)^2) = 3x? [#permalink]
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