Which of the following is always equal to sqrt(9+x^2-6x)? : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 01:14

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Which of the following is always equal to sqrt(9+x^2-6x)?

Author Message
TAGS:

### Hide Tags

Director
Joined: 09 Aug 2006
Posts: 763
Followers: 1

Kudos [?]: 192 [2] , given: 0

Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

### Show Tags

22 Oct 2007, 22:02
2
KUDOS
21
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

58% (01:39) correct 42% (00:41) wrong based on 993 sessions

### HideShow timer Statistics

Which of the following is always equal to $$\sqrt{9+x^2-6x}$$?

A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Feb 2013, 03:34, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
VP
Joined: 10 Jun 2007
Posts: 1459
Followers: 7

Kudos [?]: 255 [6] , given: 0

### Show Tags

23 Oct 2007, 10:53
6
KUDOS
2
This post was
BOOKMARKED
|x-3|

Don't for get that...

|x-3| = |3-x|

Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93298 [2] , given: 10555

Re: Which of the following is always equal to sqrt (9+x^2-6x)? [#permalink]

### Show Tags

19 Feb 2013, 04:09
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
Sachin9 wrote:
This is true and absolute value properties confirms this because:
|a-b| = |b-a|

Bunuel/KArishma,
Is this always true?

Yes, since both |a-b| and |b-a| represent the distance between a and b on the number line.

COMPLETE SOLUTION:

Which of the following is always equal to $$\sqrt{9+x^2-6x}$$?
A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93298 [2] , given: 10555

Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

### Show Tags

07 Apr 2015, 09:48
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
ankittiss wrote:
According to my understanding there are two cases in the GMAT:
Case 1 :

sqrt[(-5)*(-5)] = -5

Case 2 :

sqrt[(5)*(5)] = 5

Is my understanding correct?

Ankit

No, it's totally wrong.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

About $$\sqrt{x^2}=|x|$$:

Again, the point here is that since square root function can not give negative result then $$\sqrt{some \ expression}\geq{0}$$.

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: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.

Hope it helps.
_________________
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [1] , given: 0

### Show Tags

23 Oct 2007, 13:00
1
KUDOS
pmenon wrote:
Dont forget that square root produces a positive and negative result.

I end up with +/- (x-3). In absolute value terms, this would be |x-3| .... I dont see this as an answer choice though :-S

Sorry, but this statment is wrong... Sqrt( "Something" ) >= 0... It's never negative. In addition, "Somtehing" must be positive or 0.

Director
Joined: 09 Aug 2006
Posts: 763
Followers: 1

Kudos [?]: 192 [1] , given: 0

### Show Tags

23 Oct 2007, 20:38
1
KUDOS
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 181

Kudos [?]: 1965 [1] , given: 219

Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

### Show Tags

15 Jun 2013, 09:02
1
KUDOS
WholeLottaLove wrote:
I originally said (x-3) was the right answer.

In essence, this is a "square root of a square" problem, is it not? In that case, isn't the result always a positive number?

Whenever you have an expression in the form $$\sqrt{x^2}$$ it becomes $$|x|$$.
So in this case $$\sqrt{(x-3)^2}=|x-3|$$

For example if $$\sqrt{x^2}=3$$ x could be 3 and $$\sqrt{3^2}=3$$
but could also be -3 as $$\sqrt{(-3)^2}=3$$.

That's why we need the abs value $$x=|3|$$
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

22 Oct 2007, 22:04
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 147 [0], given: 2

### Show Tags

23 Oct 2007, 09:46
1
This post was
BOOKMARKED
Dont forget that square root produces a positive and negative result.

I end up with +/- (x-3). In absolute value terms, this would be |x-3| .... I dont see this as an answer choice though :-S
Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 190 [0], given: 7

### Show Tags

23 Oct 2007, 11:02
1
This post was
BOOKMARKED
GK_Gmat wrote:
Which of the following is always equal to sqrt (9+x^2-6x)?

a) x - 3
b) 3 + x
c) |3 - x|
d) |3 + x|
e) 3 - x

Pls. explain.

C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.
Manager
Joined: 02 Feb 2007
Posts: 120
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

23 Oct 2007, 11:32
So in other words sq rt of (x-3)^2 is +ve or -ve x-3? Sorry, but i don't really get that. I thought when we have a square of a sq rt, the answer is just what's under the sq rt, without any power.

It's not sq rt of 9 to be +-3, it's sq rt of 9^2, which is 9.
Please correct me if i am wrong.
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

23 Oct 2007, 13:05
Fistail wrote:
GK_Gmat wrote:
Which of the following is always equal to sqrt (9+x^2-6x)?

a) x - 3
b) 3 + x
c) |3 - x|
d) |3 + x|
e) 3 - x

Pls. explain.

C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.

Sorry as well... the reasonning is wrong

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression
Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 190 [0], given: 7

### Show Tags

23 Oct 2007, 13:30
Fig wrote:
Fistail wrote:
GK_Gmat wrote:
Which of the following is always equal to sqrt (9+x^2-6x)?

a) x - 3
b) 3 + x
c) |3 - x|
d) |3 + x|
e) 3 - x

Pls. explain.

C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.

Sorry as well... the reasonning is wrong

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression

i did not get your point.

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

23 Oct 2007, 13:31
Fistail wrote:
Fig wrote:
Fistail wrote:
GK_Gmat wrote:
Which of the following is always equal to sqrt (9+x^2-6x)?

a) x - 3
b) 3 + x
c) |3 - x|
d) |3 + x|
e) 3 - x

Pls. explain.

C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.

Sorry as well... the reasonning is wrong

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression

i did not get your point.

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.

Yes... Both left sides in bold are ok... but the right sides are not ... They could be negative...
Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 190 [0], given: 7

### Show Tags

23 Oct 2007, 20:05
Fig wrote:
Fistail wrote:
Fig wrote:
Fistail wrote:
GK_Gmat wrote:
Which of the following is always equal to sqrt (9+x^2-6x)?

a) x - 3
b) 3 + x
c) |3 - x|
d) |3 + x|
e) 3 - x

Pls. explain.

C. sqrt (9 + x^2 - 6x) can be written as = sqrt (x^2 - 6x + 9)
sqrt (9 + x^2 - 6x) = 3 - x
sqrt (x^2 - 6x + 9) = x - 3

(3-x) and (x-3) are not same but if we multiply either of them by -ve, we get the other. Therefore

sqrt (9 + x^2 - 6x) = sqrt (x^2 - 6x + 9) = l3-xl
so C should be it.

Sorry as well... the reasonning is wrong

We have no clues on the sign neither of 3-x nor of x-3.... So we cannot say sqrt( "Something" ) = Unknown sign expression

i did not get your point.

sqrt (9 + x^2 - 6x) and sqrt (x^2 - 6x + 9), both, are +ves.

Yes... Both left sides in bold are ok... but the right sides are not ... They could be negative...

but Fig i am still not clear on whether you mean "sqrt (9 + x^2 - 6x)" is only +ve or -ve too.

I say it is only +ve because "sqrt (9 + x^2 - 6x)" is a factor of "(9 + x^2 - 6x)" and the other fasctor is "-sqrt(9 + x^2 - 6x)". so

"sqrt (9 + x^2 - 6x)" can only be 3-x but it can be x-3 as well because we can re-write "sqrt (9 + x^2 - 6x)" as "sqrt (x^2 - 6x + 9)", which is x-3.

thanks.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 7

Kudos [?]: 189 [0], given: 0

### Show Tags

23 Oct 2007, 21:04
GK_Gmat wrote:
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 190 [0], given: 7

### Show Tags

23 Oct 2007, 21:45
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 7

Kudos [?]: 189 [0], given: 0

### Show Tags

24 Oct 2007, 00:31
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

the unknown x in sqrt(x^2) can be negative - the outcome will be positive ---> sqrt(-2^2) = 2 ---> x=-2

SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

24 Oct 2007, 02:27
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

No... ... In bold, we can only say : sqrt(x^2) = |x|

KillerSquirrel gave an exemple that proves it as well
Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 190 [0], given: 7

### Show Tags

24 Oct 2007, 05:38
Fig wrote:
Fistail wrote:
GK_Gmat wrote:
KillerSquirrel wrote:
GK_Gmat wrote:
Fig wrote:
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|

Fig can you pls. explain:

sqrt (9 + x^2 - 6x) = sqrt (x-3)^2

From this how do you arrive at :

= sqrt( (3-x)^2 )
= |3 - x|

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

No... ... In bold, we can only say : sqrt(x^2) = |x|

KillerSquirrel gave an exemple that proves it as well

Fig, I am still not convinced. I am in line with what Honghu and hobbit said here:

http://www.gmatclub.com/forum/t40319?po ... c&start=20

for me, sqrt (9+x^2-6x) = 3-x
24 Oct 2007, 05:38

Go to page    1   2    Next  [ 38 posts ]

Similar topics Replies Last post
Similar
Topics:
2 14! is equal to which of the following? 9 20 Apr 2016, 00:04
1 Which of the following procedures is always equivalent to adding 5 2 26 Nov 2015, 04:44
2 Which of the following is equal to 8 28 Jun 2013, 22:12
11 Which of the following inequalities is always true for any 9 09 Nov 2012, 10:03
4 Which of the following is always equal to sqrt(9+x^2-6x)? 6 10 Aug 2010, 22:20
Display posts from previous: Sort by