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What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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Updated on: 03 Jun 2015, 02:48
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What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined? A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\) Source: OptimusPrep
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Originally posted by naeln on 02 Jun 2015, 23:43.
Last edited by Bunuel on 03 Jun 2015, 02:48, edited 1 time in total.
Renamed the topic and edited the question.




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What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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03 Jun 2015, 02:55
naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and thus x  3 = (x  3). STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E.
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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03 Jun 2015, 03:15
Bunuel wrote: naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and x  3 = (x  3). STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E. Thank you Bunuel for the clear explanation. Love it



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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03 Jun 2015, 03:55
√x is defined means x is positive hence x is negative. For the first term if we simplify top term it becomes √(x3)^2 which is equal to x3 or 3x. Since only positive quantity comes out of sq root 3x, which is positive comes out and cancels with denominator to give 1. Kudos



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What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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03 Jun 2015, 06:06
naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep Hi Readers, At first, There has to be an observation to be madeSince each term is correctly defined in the expression \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) therefore we must realize that \(\sqrt{x}\) should be a real number which means that x MUST be a Negative NumberTherefore, (x3) MUST be Negative therefore, (x3) = \(\sqrt{(x3)^2}\) = \(\sqrt{x^2 6x+9}\) Now, Expression, \(\frac{\sqrt{x^2 6x+9}}{x3}\) can be written here as \(\frac{\sqrt{x^2 6x+9}}{\sqrt{x^2 6x+9}}\) i.e. equal to (1) Hence the result becomes \(1+\sqrt{x}\) Answer: Option
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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03 Jun 2015, 06:18
naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep ALTERNATE METHOD:Hi Readers, SUBSTITUTION of a correct numbers works very well in such cases but there has to be an observation to be madeSince each term is correctly defined in the expression \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) therefore we must realize that \(\sqrt{x}\) should be a real number which means that x MUST be a Negative NumberSo let's take X = 4 [I am not taking X = 1 because that may confuse us among options] Now, \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) = \(\frac{\sqrt{(4)^2 6(4)+9}}{(4)3}+\sqrt{(4)}\) = \(\frac{\sqrt{16 + 24+9}}{43}+\sqrt{(4)}\) = \(\frac{\sqrt{49}}{7}+2\) = \(\frac{7}{7}+2\) = \(1+2\) = \(1\) Check Options By Substituting \(X=4\) A. \(\sqrt{x}\) = 2 INCORRECTB. \(1 + \sqrt{x}\) = NOT REAL NUMBER therefore INCORRECTC. \(1 + \sqrt{x}\) = \(1+2\) = \(3\) INCORRECTD. \(1 + \sqrt{x}\) = NOT REAL NUMBER therefore INCORRECTE. \(1 + \sqrt{x}\) = \(1 + 2\) = \(+1\) CORRECTAnswer: Option
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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18 Dec 2015, 18:34
i thought it is way to easy to be 1+sqrt(x) if we have sqrt(x), it must be that x is either 0 or negative. if x is negative, then x3 is negative, thus x3 = (x3) or 1(x3). now, 1+sqrt(x) is E, which should be the correct answer.



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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21 May 2016, 16:34
Bunuel wrote: naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and thus x  3 = (x  3). STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E. Hi Brunel  why wouldn't the answer be D, since X <= 0, meaning why wouldn't sqrt(x) become sqrt(x)?



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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21 May 2016, 20:33
mdacosta wrote: Bunuel wrote: naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and thus x  3 = (x  3). STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E. Hi Brunel  why wouldn't the answer be D, since X <= 0, meaning why wouldn't sqrt(x) become sqrt(x)? Hi, D is \(1 + \sqrt{x}\) so \(\sqrt{x}\) means square root of some negative integers, and this is not defined in GMAT so this is out what you are talking of is present in E E. \(1 + \sqrt{x}\) \(\sqrt{x}\).. since x<=0, the square root turns into that of +ive number.. so E is correct
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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20 Jun 2016, 06:59
Hi Bunuel, I have completed the Manhattan GMAT Algebra Guide and it says that sqrt(y), where y>=0, is positive only; while for the expression x^2= y, where y>=0, is mod(y), i.e. y can be positive or negative. Now you have solved the first term of this expression by considering both positive and negative values. This is confusing me. Please, provide some insight. Thanks in advance.
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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20 Jun 2016, 07:07
oloz wrote: Hi Bunuel, I have completed the Manhattan GMAT Algebra Guide and it says that sqrt(y), where y>=0, is positive only; while for the expression x^2= y, where y>=0, is mod(y), i.e. y can be positive or negative. Now you have solved the first term of this expression by considering both positive and negative values. This is confusing me. Please, provide some insight. Thanks in advance. That's not true. We got that \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). x3, on the other hand due to the fact that \(x\leq{0}\), equals to (x  3), which is positive.
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What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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21 Jun 2016, 05:40
Bunuel wrote: naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and thus x  3 = (x  3).STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E. I got this explanation fully except for the underlined part. x is negative, but x3should still be positive. I dont understand how we get (x3) from x3. Can someone please help me understand this part? 'll be much obliged. Thanks.



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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26 Sep 2017, 20:55
iqbalfiery wrote: Bunuel wrote: naeln wrote: What is \(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}\) if each term in this expression is correctly defined?
A. \(\sqrt{x}\) B. \(1 + \sqrt{x}\) C. \(1 + \sqrt{x}\) D. \(1 + \sqrt{x}\) E. \(1 + \sqrt{x}\)
Source: OptimusPrep STEP 1:Even roots from a negative numbers are not defined, thus \(\sqrt{x}\) to be defined x must be more than or equal to 0: \(x\geq{0}\). Therefore \(x\leq{0}\). STEP 2:Recall that \(\sqrt{x^2}=x\). \(\sqrt{x^2 6x+9}=\sqrt{(x3)^2}=x3\). Since from above \(x\leq{0}\), then x3 is negative and thus x  3 = (x  3).STEP 3:\(\frac{\sqrt{x^2 6x+9}}{x3}+\sqrt{x}=\frac{(x3)}{x3} +\sqrt{x}=1+\sqrt{x}\). Answer: E. I got this explanation fully except for the underlined part. x is negative, but x3should still be positive. I dont understand how we get (x3) from x3. Can someone please help me understand this part? 'll be much obliged. Thanks. Modulus is always positive ie Ix3I has to be positive and is also=(x3) as per above. Now we have also derived that x<=0 ie negative or 0. on a number line, we have ........1......0......+1........ let us say x is 1, therefore, we have (13)=4 which is equal to Ix3I which is not possible since modulus has to be +ve. Therefore we put sign in the front of (x3) to make it positive ie (13)=(4)=+4 This also works if we take x=0.



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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02 Oct 2017, 00:46
BunuelIsn't that if something is under even root on gmat, we only consider the positive solution?



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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02 Oct 2017, 00:48
kaleem765 wrote: BunuelIsn't that if something is under even root on gmat, we only consider the positive solution? Yes. Even roots from negative numbers are not defined on the GMAT.
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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02 Oct 2017, 00:53
Bunuel wrote: kaleem765 wrote: BunuelIsn't that if something is under even root on gmat, we only consider the positive solution? Yes. Even roots from negative numbers are not defined on the GMAT. Then why did we consider (x3) as a solution for \sqrt{x^26x+9}?



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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02 Oct 2017, 00:59
kaleem765 wrote: Bunuel wrote: kaleem765 wrote: BunuelIsn't that if something is under even root on gmat, we only consider the positive solution? Yes. Even roots from negative numbers are not defined on the GMAT. Then why did we consider (x3) as a solution for \sqrt{x^26x+9}? Since we got that x must be less than or equal to 0, then (x  3) is positive not negative.
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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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02 Oct 2017, 01:04
Thanks Bunuel. I got it now.



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Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
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31 Oct 2017, 03:06
I'm sorry  I still don't get it. Quant is definitely my weak point, but I simply cannot understand how we get to (x3)... I'd hugely appreciate anyone spelling it out for me in layman's terms (as much as that is possible)  or pointing me to a relevant article where I can get into it there.
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