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

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

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22 Oct 2007, 22:02
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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.
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23 Oct 2007, 10:53
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|x-3|

Don't for get that...

|x-3| = |3-x|

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

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19 Feb 2013, 04:09
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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

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

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

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07 Apr 2015, 09:48
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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.
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23 Oct 2007, 13:00
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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.

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23 Oct 2007, 20:38
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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.
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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]

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15 Jun 2013, 09:02
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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 $$x=|3|$$
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22 Oct 2007, 22:04
(C) for me

sqrt (9+x^2-6x)
= sqrt( (3-x)^2 )
= |3-x|
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23 Oct 2007, 09:46
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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
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23 Oct 2007, 11:02
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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.
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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.
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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
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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.
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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...
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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.
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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.

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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.
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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

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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
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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

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