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

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

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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, 04:34, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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The answer is indeed
|x-3|

Don't for get that...

|x-3| = |3-x|

C is the answer.

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

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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

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

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

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ankittiss wrote:
Bunuel Can you please help here?
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?

Thanks in advance
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 \(x=\sqrt{9}=3\).

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

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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


There should be no confusion regarding this question.
SQUAREROOT and MODULUS have the same property regarding the polarity of a number. THEY BOTH YIELDS ONLY POSITIVE OUTPUTS.

Q:- what is the surest way to make any value positive. ?
A:- Take the modulus of the value |x|

So we know \(\sqrt{9+x^2-6x}\) will gives us ONLY POSITIVE VALUE, Then we should make sure that the option also matches this property.
Therefore only |3 - x| and |3+x| are the one that will always give positive value
BUT |3+x| is not a root or solution of \(\sqrt{9+x^2-6x}\)
So the only option left is |3-x|

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

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Madhavi1990 wrote:
After factorization I got sq root x- 3 ^2 => (x-3) ^2 so x = 3
I tried this algebraically - keeping in mind 'always true'. I would be glad to hear any comments on the same
1) x - 3 = 0 can be 3 => 3 - 3 = or x = 3 but I wasn't sure if this will hold always
2) x = -3 (not correct)
3) |3-x|: 3 - x =0; so x=3 and -3+x=0 so x = 3 (here both negative or positive sign; value remains the same)
4) |3+x| = 0 will not hold
5) 3 - x= 0; so -x = - 3 or x = 3

Now I know, that C will ALWAYS be true. But Why is A, E wrong? They are giving us 3 also
Can anyone explain the logic for A, E to be wrong?



Hey as per the expression:

9+x2−6x‾‾‾‾‾‾‾‾‾‾‾√=(3−x)2‾‾‾‾‾‾‾‾√=|3−x|9+x2−6x=(3−x)2=|3−x|.

You can't equate it to 0 and compute value of x as 3. Anyways A and E are subset of the solution. In case of C it is considering both the solutions which equate to square root expression given in question stem.

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New post 22 Oct 2007, 23:04
(C) for me :)

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

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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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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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New post 23 Oct 2007, 12:32
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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New post 23 Oct 2007, 14:05
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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New post 23 Oct 2007, 14:30
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. :roll:

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

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New post 23 Oct 2007, 14:31
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. :roll:

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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New post 23 Oct 2007, 21:05
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. :roll:

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.

please correct me and also clearify your reasoning..

thanks.

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New post 23 Oct 2007, 22:04
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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New post 23 Oct 2007, 22:45
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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New post 24 Oct 2007, 01:31
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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