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Which of the following options always equates to \(\sqrt{9 + x^2 - 6x}\)?
A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x
A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x
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Sachin9
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 options always equates to \(\sqrt{9 + x^2 - 6x}\)?
A. \(x - 3\)
B. \(3 + x\)
C. \(|3 - x|\)
D. \(|3 + x|\)
E. \(3 - x\)
\(\sqrt{9 + x^2 - 6x} = \)
\(=\sqrt{(3 - x)^2} = \)
\(=|3 - x|\)
Note that both \(|x - 3|\) and \(|3 - x|\) represent the distance between \(x\) and 3. Hence, it follows that \(|x - 3| = |3 - x|\).
Answer: C
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ankittiss
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
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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General Discussion
The answer is indeed
|x-3|
Don't for get that...
|x-3| = |3-x|
C is the answer.
|x-3|
Don't for get that...
|x-3| = |3-x|
C is the answer.
This is true and absolute value properties confirms this because:
|a-b| = |b-a|
Bunuel/KArishma,
Is this always true?
|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
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
mehasingh
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.
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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
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
GMAT 1: 750 Q49 V43
Posts: 463
GK_Gmat
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
Posts: 8
Isn't it ok if (x - 3) is negative though because you are squaring it prior to taking the square root? It will never be the square root of a negative.
Bunuel
sixfivefive
\(\sqrt{9 + x^2 - 6x} = \)
\(=\sqrt{(3 - x)^2} = \)
\(=|3 - x|=|x-3|\)
If x - 3 is negative then |x - 3| = -(x - 3) = 3 - x, which in this case will be positive. So, in any case we get posiitve result.
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Posts: 477
GK_Gmat
Method 1: Algebraic Approach
The expression inside the square root is a perfect square trinomial.
Rearrange the terms:
\(\sqrt{9 + x^2 - 6x} = \sqrt{x^2 - 6x + 9}\)
Factor the quadratic expression:
\(x^2 - 6x + 9 = (x - 3)^2\) (or equivalently \((3 - x)^2\))
So the expression becomes:
\(\sqrt{(x - 3)^2}\)
CRITICAL RULE:
In GMAT math, \(\sqrt{u^2}\) is defined as \(|u|\), not just \(u\). The square root symbol \(\sqrt{}\) only yields the non-negative root.
Therefore:
\(\sqrt{(x - 3)^2} = |x - 3|\)
Note that \(|x - 3|\) is mathematically identical to \(|3 - x|\).
If you check the options, \(|3 - x|\) is Option (C).
[hr]
Method 2: Picking Numbers
If you forget the absolute value rule, picking numbers will save you.
Case 1: Pick \(x = 0\) (Choose a value smaller than 3)
Substitute into the original expression:
\(\sqrt{9 + 0^2 - 6(0)} = \sqrt{9} = 3\)
Now check the options:
A. \(0 - 3 = -3\) (Incorrect)
B. \(3 + 0 = 3\) (Keep)
C. \(|3 - 0| = 3\) (Keep)
D. \(|3 + 0| = 3\) (Keep)
E. \(3 - 0 = 3\) (Keep)
Case 2: Pick \(x = 5\) (Choose a value larger than 3)
Substitute into the original expression:
\(\sqrt{9 + 25 - 30} = \sqrt{4} = 2\)
Now check the remaining options:
B. \(3 + 5 = 8\) (Incorrect)
C. \(|3 - 5| = |-2| = 2\) (Correct)
D. \(|3 + 5| = 8\) (Incorrect)
E. \(3 - 5 = -2\) (Incorrect, remember the result of a square root must be positive)
Answer: C
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