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

This is true and absolute value properties confirms this because:
|a-b| = |b-a|

Bunuel/KArishma,
Is this always true?

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


Hi, can anyone explain me how to go abt this...the above answers have got a
lil confusing for me.

ill go one option at a time:

A. x-3

when we square x-3 it give x^2+9-6x
if we squareroot x^2+9-6x then we get the same exp...so for me this seems
to be an answer

B. 3+x

when we square 3+x it gives 9+6x+x^2
this is all positive unlike 9+x^2-6x given in the question, so not an answer

C. |3-x|

given sqrt{9+x^2-6x}

solving sqrt(3-x)

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

this too seems to be a possible answer

D. |3+x|

this will give 9+x^2+6x which is not equal to the equation given hence not
an answer

E. 3-x

=9+x^2-6x

this too seems fine

can anyone please clarify where im going wrong.... in all the options...


Thanks

Notice that the square root function cannot give negative result: \(\sqrt{{some \ expression}}\geq{0}\).

So, \(\sqrt{9+x^2-6x}=\sqrt{(3-x)^2}\geq{0}\).

Now, in option A we have x-3, which can be negative if x<3, so A cannot be the correct answer.

Hope it's clear.

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 \(x=\sqrt{9}=3\).



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.

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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