GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 08 Dec 2019, 03:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 05 Aug 2014
Posts: 27
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
5
55
00:00

Difficulty:

95% (hard)

Question Stats:

36% (01:36) correct 64% (01:16) wrong based on 872 sessions

### HideShow timer Statistics

What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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

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.
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 59592
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
10
25
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Answer: E.
_________________
##### General Discussion
Intern
Joined: 05 Aug 2014
Posts: 27
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 $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Answer: E.

Thank you Bunuel for the clear explanation. Love it
Intern
Joined: 12 May 2014
Posts: 13
Location: United States
Concentration: Strategy, Operations
Schools: IIMC'17
GMAT Date: 10-22-2014
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 √(x-3)^2 which is equal to x-3 or 3-x. Since only positive quantity comes out of sq root 3-x, which is positive comes out and cancels with denominator to give -1.

Kudos
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
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
1
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 made

Since each term is correctly defined in the expression
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$

therefore we must realize that $$\sqrt{-x}$$ should be a real number which means that x MUST be a Negative Number

Therefore, (x-3) MUST be Negative

therefore, (x-3) = $$-\sqrt{(x-3)^2}$$ = $$-\sqrt{x^2 -6x+9}$$

Now, Expression, $$\frac{\sqrt{x^2 -6x+9}}{x-3}$$ 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
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One 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
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
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 $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 made

Since each term is correctly defined in the expression
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$

therefore we must realize that $$\sqrt{-x}$$ should be a real number which means that x MUST be a Negative Number

So let's take X = -4 [I am not taking X = -1 because that may confuse us among options]

Now, $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ = $$\frac{\sqrt{(-4)^2 -6(-4)+9}}{(-4)-3}+\sqrt{-(-4)}$$

= $$\frac{\sqrt{16 + 24+9}}{-4-3}+\sqrt{(4)}$$
= $$\frac{\sqrt{49}}{-7}+2$$
= $$\frac{7}{-7}+2$$
= $$-1+2$$
= $$1$$

Check Options By Substituting $$X=-4$$

A. $$\sqrt{-x}$$ = 2 INCORRECT
B. $$1 + \sqrt{x}$$ = NOT REAL NUMBER therefore INCORRECT
C. $$1 + \sqrt{-x}$$ = $$1+2$$ = $$3$$ INCORRECT
D. $$-1 + \sqrt{x}$$ = NOT REAL NUMBER therefore INCORRECT
E. $$-1 + \sqrt{-x}$$ = $$-1 + 2$$ = $$+1$$ CORRECT

Answer: Option
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One 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: 2492
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
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
1
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 x-3 is negative, thus |x-3| = -(x-3) or -1(x-3).
now, -1+sqrt(-x) is E, which should be the correct answer.
Manager
Joined: 05 Dec 2015
Posts: 96
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 $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

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: 8288
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
1
mdacosta wrote:
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

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
_________________
Intern
Status: It's time...
Joined: 29 Jun 2015
Posts: 7
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q48 V25
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: 59592
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 $$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. |x-3|, on the other hand due to the fact that $$x\leq{0}$$, equals to -(x - 3), which is positive.
_________________
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 $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Answer: E.

I got this explanation fully except for the underlined part. x is negative, but |x-3|should still be positive. I dont understand how we get -(x-3) from |x-3|. Can someone please help me understand this part? 'll be much obliged. Thanks.
Manager
Joined: 04 May 2014
Posts: 150
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 $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ 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 $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Answer: E.

I got this explanation fully except for the underlined part. x is negative, but |x-3|should still be positive. I dont understand how we get -(x-3) from |x-3|. Can someone please help me understand this part? 'll be much obliged. Thanks.

Modulus is always positive ie Ix-3I has to be positive and is also=(x-3) 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 (-1-3)=-4 which is equal to Ix-3I which is not possible since modulus has to be +ve.
Therefore we put -sign in the front of (x-3) to make it positive ie -(-1-3)=-(-4)=+4
This also works if we take x=0.
Intern
Joined: 24 Oct 2016
Posts: 24
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
1
Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?
Math Expert
Joined: 02 Sep 2009
Posts: 59592
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
kaleem765 wrote:
Bunuel
Isn'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.
_________________
Intern
Joined: 24 Oct 2016
Posts: 24
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
1
Bunuel wrote:
kaleem765 wrote:
Bunuel
Isn'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 -(x-3) as a solution for \sqrt{x^2-6x+9}?
Math Expert
Joined: 02 Sep 2009
Posts: 59592
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
kaleem765 wrote:
Bunuel wrote:
kaleem765 wrote:
Bunuel
Isn'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 -(x-3) as a solution for \sqrt{x^2-6x+9}?

Since we got that x must be less than or equal to 0, then -(x - 3) is positive not negative.
_________________
Intern
Joined: 24 Oct 2016
Posts: 24
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
GMAT 1: 740 Q49 V41
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 -(x-3)... 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!
Non-Human User
Joined: 09 Sep 2013
Posts: 13724
Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this  [#permalink]

### Show Tags

21 Nov 2018, 00:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this   [#permalink] 21 Nov 2018, 00:30
Display posts from previous: Sort by

# What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne