Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 05 Aug 2014
Posts: 30

What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
Updated on: 03 Jun 2015, 02:48
Question Stats:
36% (01:37) correct 64% (01:14) wrong based on 696 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
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.






Math Expert
Joined: 02 Sep 2009
Posts: 50003

What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
03 Jun 2015, 02:55




Intern
Joined: 05 Aug 2014
Posts: 30

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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



Intern
Joined: 12 May 2014
Posts: 14
Location: United States
Concentration: Strategy, Operations
GMAT Date: 10222014
GPA: 1.9
WE: Engineering (Energy and Utilities)

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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



CEO
Joined: 08 Jul 2010
Posts: 2559
Location: India
GMAT: INSIGHT
WE: Education (Education)

What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



CEO
Joined: 08 Jul 2010
Posts: 2559
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Board of Directors
Joined: 17 Jul 2014
Posts: 2657
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.



Manager
Joined: 05 Dec 2015
Posts: 118

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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)?



Math Expert
Joined: 02 Aug 2009
Posts: 6961

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Intern
Status: It's time...
Joined: 29 Jun 2015
Posts: 7
Location: India
Concentration: Operations, General Management
GPA: 3.59

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.
_________________
Perseverance, the answer to all our woes. Amen.



Math Expert
Joined: 02 Sep 2009
Posts: 50003

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 21 Feb 2016
Posts: 9
Location: United States (MA)

What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.



Manager
Joined: 04 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.



Intern
Joined: 24 Oct 2016
Posts: 25

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



Math Expert
Joined: 02 Sep 2009
Posts: 50003

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
02 Oct 2017, 00:48



Intern
Joined: 24 Oct 2016
Posts: 25

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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}?



Math Expert
Joined: 02 Sep 2009
Posts: 50003

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
02 Oct 2017, 00:59



Intern
Joined: 24 Oct 2016
Posts: 25

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
02 Oct 2017, 01:04
Thanks Bunuel. I got it now.



Intern
Joined: 30 May 2017
Posts: 12
GPA: 3.9

Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this
[#permalink]
Show Tags
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.
Many thanks!




Re: What is (x^2  6x + 9)^(1/2)/(x  3) + (x)^(1/2) if each term in this &nbs
[#permalink]
31 Oct 2017, 03:06






