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# M19-20

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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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16 Sep 2014, 00:06
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Difficulty:

35% (medium)

Question Stats:

60% (00:30) correct 40% (00:34) wrong based on 209 sessions

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

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

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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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16 Sep 2014, 00:06
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Official Solution:

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

A. $$x - 3$$
B. $$3 + x$$
C. $$|3 - x|$$
D. $$|3 + x|$$
E. $$3 - x$$

$$\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|$$ (by the definition of a square root).

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Joined: 04 Jul 2014
Posts: 306
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
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WE: Analyst (Accounting)

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21 Nov 2014, 02:18
1
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Hi Bunuel,

I got the roots for the equation as (x-3)^2 (reading the equation as x^2 - 6x +9 - We've flipped the values of a and b in the equation a^2 - 2ab + b^2). Based on my approach |x - 3| would be the correct answer and on yours |3 - x| is the correct answer.

Does my above para make any sense? If it does, and if the question has both of these options, which would be the correct answer? If it doesn't please help me understand my mistake.
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Cheers!!

JA
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Math Expert
Joined: 02 Sep 2009
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21 Nov 2014, 03:51
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joseph0alexander wrote:
Hi Bunuel,

I got the roots for the equation as (x-3)^2 (reading the equation as x^2 - 6x +9 - We've flipped the values of a and b in the equation a^2 - 2ab + b^2). Based on my approach |x - 3| would be the correct answer and on yours |3 - x| is the correct answer.

Does my above para make any sense? If it does, and if the question has both of these options, which would be the correct answer? If it doesn't please help me understand my mistake.

The point is that |x - 3| = |3 - x|. Both indicate the distance between x and 3.
_________________
Current Student
Joined: 04 Jul 2014
Posts: 306
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)

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21 Nov 2014, 04:05
1
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Bunuel wrote:
The point is that |x - 3| = |3 - x|. Both indicate the distance between x and 3.

So, I understand that both of our answers are correct and that both these values won't appear as options in a single question.
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Cheers!!

JA
If you like my post, let me know. Give me a kudos!

Math Expert
Joined: 02 Sep 2009
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21 Nov 2014, 04:07
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Expert's post
joseph0alexander wrote:
Bunuel wrote:
The point is that |x - 3| = |3 - x|. Both indicate the distance between x and 3.

So, I understand that both of our answers are correct and that both these values won't appear as options in a single question.

____________
Yes, that's correct.
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Joined: 14 May 2014
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GMAT 1: 700 Q44 V41
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24 Jul 2015, 06:44
Bunuel wrote:
Official Solution:

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

A. $$x - 3$$
B. $$3 + x$$
C. $$|3 - x|$$
D. $$|3 + x|$$
E. $$3 - x$$

$$\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|$$ (by the definition of a square root).

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help
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Joined: 02 Sep 2009
Posts: 43853

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24 Jul 2015, 06:51
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riyazgilani wrote:
Bunuel wrote:
Official Solution:

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

A. $$x - 3$$
B. $$3 + x$$
C. $$|3 - x|$$
D. $$|3 + x|$$
E. $$3 - x$$

$$\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|$$ (by the definition of a square root).

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help

Exactly because the square root function cannot give negative result the answer is |3-x|, an absolute value of 3-x, which also cannot be negative.

MUST KNOW: $$\sqrt{x^2}=|x|$$:

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 $$\sqrt{x^2}=|x|$$.

I'd advice to go through basics and only after to practice questions:

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Joined: 03 Dec 2013
Posts: 72
Location: United States (HI)
GMAT 1: 660 Q49 V30
GPA: 3.56

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22 Aug 2016, 13:02
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 18 Jun 2015
Posts: 41

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12 Sep 2016, 14:23
1
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After solving this question I got the result as |X-3| and I answered it as X-3, And got this wrong.
Thanks for giving the link which clears the basic of Absolute value including the result:

|A-B| = |B-A|

Which is clearly applicable in this question provided options.
Manager
Joined: 14 Oct 2012
Posts: 179

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30 Mar 2017, 19:19
1
KUDOS
Bunuel wrote:
riyazgilani wrote:
Bunuel wrote:
Official Solution:

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

A. $$x - 3$$
B. $$3 + x$$
C. $$|3 - x|$$
D. $$|3 + x|$$
E. $$3 - x$$

$$\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|$$ (by the definition of a square root).

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help

Exactly because the square root function cannot give negative result the answer is |3-x|, an absolute value of 3-x, which also cannot be negative.

MUST KNOW: $$\sqrt{x^2}=|x|$$:

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 $$\sqrt{x^2}=|x|$$.

I'd advice to go through basics and only after to practice questions:

Hello Bunuel, Vyshak
I thought the equation was as follows:
$$\sqrt{x^2}=x$$, if $$x>{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<={0}$$.
Am i wrong? Please correct me if so.
Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 43853

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30 Mar 2017, 21:01
manishtank1988 wrote:
Hello Bunuel, Vyshak
I thought the equation was as follows:
$$\sqrt{x^2}=x$$, if $$x>{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<={0}$$.
Am i wrong? Please correct me if so.
Thanks

You can put = sign in any of the two because $$\sqrt{0^2}=0=-0$$.
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Joined: 26 Mar 2013
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31 Mar 2017, 09:00
Dear Bunuel,

The answer choices CAN"T have the following answers in their list: $$|3 - x|$$ and $$|x - 3|$$ Because the following rule:

$$|3 - x|$$ = $$|x - 3|$$

Am I right?

Manager
Joined: 14 Oct 2012
Posts: 179

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31 Mar 2017, 14:07
Bunuel wrote:
manishtank1988 wrote:
Hello Bunuel, Vyshak
I thought the equation was as follows:
$$\sqrt{x^2}=x$$, if $$x>{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<={0}$$.
Am i wrong? Please correct me if so.
Thanks

You can put = sign in any of the two because $$\sqrt{0^2}=0=-0$$.

Understood thanks...
Manager
Joined: 09 Nov 2016
Posts: 54

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21 Aug 2017, 04:28
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This question is based on the principle of $$\sqrt{x^2}=|x|$$
Hence its $$|x-3|=|3-x|$$
Straight away C.

Press Kudos id this helps!
Intern
Joined: 13 Oct 2017
Posts: 28

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09 Feb 2018, 06:42
Hi Bunuel,

I used the plug-in method...but still a little confused as to why C was correct...I did the following:

made x=2...then...√9+4-12 = √1 as my goal.

I then plugged in the two in place of x in the answer choices...to see if I got a figure that corresponded with my goal of √1.

A) 2-3= -1 NO
B) 3+2= 5 NO
C) [3-2] = [1] MAYBE
D) [3+2] = [5] NO
E) 3-2 = 1 MAYBE

I originally chose E because the goal of √1, seems to result in 1 and 1 only when one finds the square root.
However, upon a second look...is [1] another way of saying √1, thus making C the correct answer?
Math Expert
Joined: 02 Sep 2009
Posts: 43853

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09 Feb 2018, 06:49
ttaiwo wrote:
Hi Bunuel,

I used the plug-in method...but still a little confused as to why C was correct...I did the following:

made x=2...then...√9+4-12 = √1 as my goal.

I then plugged in the two in place of x in the answer choices...to see if I got a figure that corresponded with my goal of √1.

A) 2-3= -1 NO
B) 3+2= 5 NO
C) [3-2] = [1] MAYBE
D) [3+2] = [5] NO
E) 3-2 = 1 MAYBE

I originally chose E because the goal of √1, seems to result in 1 and 1 only when one finds the square root.
However, upon a second look...is [1] another way of saying √1, thus making C the correct answer?

C is correct because it's generally true, always true for ALL values of x.

As for your approach: for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example, check x = 4.
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09 Feb 2018, 07:46
Hi Bunuel,

Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)

A) 1, NO
B) 7, NO
C) [-1], MAYBE
D) [7], NO
E) -1, NO

So in this case, C is conclusively the case because it is always true for all values of x...

My question remains though...is √1 always [1], and is that also the case for √-1?

Thanks again,

Tosin
Math Expert
Joined: 02 Sep 2009
Posts: 43853

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09 Feb 2018, 07:51
ttaiwo wrote:
Hi Bunuel,

Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)

A) 1, NO
B) 7, NO
C) [-1], MAYBE
D) [7], NO
E) -1, NO

So in this case, C is conclusively the case because it is always true for all values of x...

My question remains though...is √1 always [1], and is that also the case for √-1?

Thanks again,

Tosin

$$\sqrt{1}=1$$ only.

$$\sqrt{-1}$$ is an imaginary number and since all numbers on the GMAT are real numbers you don't need to worry about that at all..

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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09 Feb 2018, 08:18
Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)

A) 1, NO
B) 7, NO
C) [-1], MAYBE
D) [7], NO
E) -1, NO

So in this case, C is conclusively the case because it is always true for all values of x...

My question remains though...is √1 always [1], and is that also the case for √-1?

Thanks again,

Tosin

$$\sqrt{1}=1$$ only.

$$\sqrt{-1}$$ is an imaginary number and since all numbers on the GMAT are real numbers you don't need to worry about that at all..

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 $$x=\sqrt{9}=3$$.

Thanks
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