GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Sep 2019, 10:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0

Author Message
TAGS:

### Hide Tags

Director
Joined: 23 Sep 2007
Posts: 682
Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

24 Apr 2008, 17:42
19
116
00:00

Difficulty:

55% (hard)

Question Stats:

61% (01:37) correct 39% (02:01) wrong based on 1142 sessions

### HideShow timer Statistics

Is $$\sqrt{(x-3)^2} = 3-x$$?

(1) $$x\neq{3}$$

(2) $$-x|x| > 0$$

Attachment:

fasdfasdfasdfasdf.JPG [ 30.21 KiB | Viewed 64489 times ]
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

13 Jun 2010, 03:41
48
60
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function cannot give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.
_________________
Manager
Joined: 24 Apr 2008
Posts: 142
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

24 Apr 2008, 21:45
10
3
From the que: 3-x is always >0 --> x has to be less than 3.

Option 1: X can also be >3 when ans fails so insufficient
Option 2: -x|x|>0 implies x is always < 0 which means x is less than 3 hence sufficient.

Ans : B
##### General Discussion
Director
Joined: 14 Aug 2007
Posts: 590
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

25 Apr 2008, 00:23
3
2
gmatnub wrote:
is sqrt ((x-3)^2) = 3-x?

1) x not equal to 3
2) -x|x| > 0

The oa is B, but why is A alone not enough?

given |x-3| can be equal to 3-x for x < 3,

1) X can be greater than 3
2) X is less than 0, i.e x < 3, for all x.

so B
Manhattan Prep Instructor
Joined: 28 Aug 2009
Posts: 154
Location: St. Louis, MO
Schools: Cornell (Bach. of Sci.), UCLA Anderson (MBA)

### Show Tags

28 Mar 2010, 12:38
4
4
One important thing to know about the square root of a square term:
(1) If it's a number under the square root sign, the answer is only the positive root: sqrt[(-3)^2] = sqrt[(3)^2] = sqrt[9] = 3.
(2) If there are variables under the square root sign, you must consider the fact that the squared "thing" may have been either pos or neg to begin with: sqrt[(-x)^2] = sqrt[(x)^2] = |x|. We are still looking for the positive root, but we don't know whether +x or -x is actually greater than zero--that depends on the sign of x.

So for this question: Is sqrt[(x-3)^2] = 3-x?

Rephrase to: Is |x-3| = 3-x?

If stuff in the absolute value sign is positive or zero, this becomes:
Is x-3 = 3-x?
Is 2x = 6?
Is x = 3?

If stuff in the absolute value sign is negative, this becomes:
Is -(x-3) = 3-x?
Is -x+3 = 3-x?
The answer is yes for all x, but remember this was just "all x" such that x-3 was negative, or x < 3: So we ask "Is x < 3?"

Put the two cases together for the final rephrase: "Is x =<3 ? "

(1) x is not 3, but no info on whether it is less than or greater than 3. INSUFF.
(2) -x|x| > 0.
|x| is positive, so -x would have to be positive too (pos*pos > 0, but neg*pos < 0). Thus, x is negative. If x is negative, it is definitely less than 3. The answer is definitely Yes. SUFF.

_________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Manhattan GMAT Discount | Manhattan GMAT Course Reviews | Manhattan GMAT Reviews
Intern
Joined: 04 Dec 2009
Posts: 16
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

13 Jun 2010, 02:57
1
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

SVP
Joined: 24 Jul 2011
Posts: 1886
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740
Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x  [#permalink]

### Show Tags

24 Jul 2011, 23:47
Lets disect the question stem first.

sqrt((x-3)^2) = 3 - x

Carrying out the squaring and then the square root operation, we get:

|(x-3)| = 3 - x

To solve the modulus sign, we will have two cases -
Case 1: (x-3)>=0
=> x = 3

Case 2: (x-3) < 0
=> Every value of x less than 3 satisfies the equation

Therefore sqrt((x-3)^2) = 3 - x
for all values of x<=3, but this equation is not satisfied for values of x>3

Now consider the statements of the DS.

Statement (1): x is not equal to 3
If x is not equal to 3, it could be greater than 3 or less than 3. If x is less than 3, the equation is satisfied. If x is greater than 3, the equation is not satisfied. Therefore we cannot say if the equation will be satisfied if x is not equal to 3.
Statement (1) alone is therefore insufficient to answer the question.

Next lets consider statement (2):
Statement (2): -x|x|>0
Again to remove the modulus sign make two cases.
Case 1: x>=0 => -x(x) >0 => -(x^2) > 0, which is impossible because x^2 is always >=0, so -(x^2) cannot be >0 for any values of x. Therefore to satisfy this inequality, x cannot be >0
Case 2: x<0 => -x (-x) > 0
=> x^2 > 0 , which is true for all values of x except 0.
Therefore the inequality is satisfied for all values of x<0

Therefore from statement (2) we know that x<0. And from the stem of the question (after we solved it), we know that the equation given in the stem is satisfied for all values of x<=3. Therefore the equation will hold for all values of x specified by statement (2). Therefore statement (2) alone is enough to solve the question.

_________________

Awesome Work | Honest Advise | Outstanding Results

Reach Out, Lets chat!
Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services
Intern
Joined: 30 Jan 2012
Posts: 13
Location: United States
Concentration: Entrepreneurship, Strategy
GMAT 1: 640 Q47 V32
GPA: 3.6
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

23 Sep 2013, 08:09
1
2
Hello Bunuel,

After this point..

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

Can this equation be written this way?

1)
x-3 = 3-x => x = 3

2)
-(x-3) = 3-x .. this leads to nothing

So I concluded that the question is whether x=3 and hence I chose A as answer.. but I am wrong.

What is that I doing wrong here?

Thanks
C23678

Bunuel wrote:
Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

24 Sep 2013, 07:22
c23678 wrote:
Hello Bunuel,

After this point..

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

Can this equation be written this way?

1)
x-3 = 3-x => x = 3

2)
-(x-3) = 3-x .. this leads to nothing

So I concluded that the question is whether x=3 and hence I chose A as answer.. but I am wrong.

What is that I doing wrong here?

Thanks
C23678

Bunuel wrote:
Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

$$|x-3|=-(x-3)$$ when $$x\leq{3}$$. In this case we'd have $$-(x-3)=3-x$$ --> 3=3 --> true. This means that when $$x\leq{3}$$, then the equation holds true.

Try numbers less than or equal to 3 to check.
_________________
Intern
Joined: 07 May 2014
Posts: 22
Concentration: Strategy, Technology
Schools: Anderson '17
GMAT 1: 730 Q50 V40
GPA: 3.1
WE: Analyst (Consulting)
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

27 Jul 2014, 10:29
1
Please clarify a doubt which i have in this question :

If we have a question, Is x<=5,
A. X<0
B. X<=0

In the original question, I am confused because 0, which satisfies the equation, doesn't appear in x|x| < 0. And hence the solution is incomplete.
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

27 Jul 2014, 14:50
vibsaxena wrote:
Please clarify a doubt which i have in this question :

If we have a question, Is x<=5,
A. X<0
B. X<=0

In the original question, I am confused because 0, which satisfies the equation, doesn't appear in x|x| < 0. And hence the solution is incomplete.

The original question asks whether $$x\leq{3}$$: the answer would be YES if x is 3 or less than 3. (2) says that $$x<0$$, so the answer is clearly YES.

Does this make sense?
_________________
Intern
Joined: 28 Jan 2013
Posts: 28
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

03 Sep 2014, 12:18
2
gmatnub wrote:
Is $$\sqrt{(x-3)^2} = 3-x$$?

(1) $$x\neq{3}$$
(2) -x|x| > 0

mod(x-3)=3-x?

mod(y) = -y when y is -ve => mod(x-3) can be equal to -(x-3) only when x-3 is negative i.e x-3<0 => x<3

1) x not equal to 3=> which means x can be greater than 3 or less than 3
2) -x|x|>0=> this is possible only when x is -ve i.e x<0

statement 2 gives x<0, so statement 2 alone solves the problem.
Intern
Joined: 01 Sep 2014
Posts: 5
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

23 Oct 2014, 23:40
Hi Bunuel
As a^2 = 25 has two solutions -------------------------> a=5 and a= -5
therefore a= sqrt 25 should also have two solutions-----> a=5 and a= -5

Then why do we say that square root of a positive no. is always positive?
Shouldn't sqrt 25 have two possible values +5 and -5. ?
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

24 Oct 2014, 03:21
1
1
arpitsharms wrote:
Hi Bunuel
As a^2 = 25 has two solutions -------------------------> a=5 and a= -5
therefore a= sqrt 25 should also have two solutions-----> a=5 and a= -5

Then why do we say that square root of a positive no. is always positive?
Shouldn't sqrt 25 have two possible values +5 and -5. ?

NO!

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

Hope it's clear.
_________________
Intern
Joined: 03 Jul 2014
Posts: 14
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

22 Nov 2014, 03:02
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

22 Nov 2014, 07:20
2
hanschris5 wrote:
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.

Sure. We have that $$-x|x| >0$$, so the product of two multiples -x and |x| is positive. Now, we know that |x| must be positive, hence -x must also be positive for the product to be positive, thus -x > 0, which is the same as x < 0.

Hope it's clear.
_________________
Director
Joined: 26 Oct 2016
Posts: 620
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

26 Jan 2017, 19:55
1
Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.
The DISTANCE between two numbers must be greater than or equal to 0.

The question stem above, rephrased: Is |x-3| = 3-x?
In words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE of 3 and x?
The answer will be YES if the DIFFERENCE of 3 and x is greater than or equal to 0:
3-x≥0
x≤3.

The question stem rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be positive*positive or negative*negative.
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

_________________
Thanks & Regards,
Anaira Mitch
Intern
Joined: 21 Oct 2017
Posts: 2
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

02 Oct 2018, 11:30
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.
Math Expert
Joined: 02 Sep 2009
Posts: 58092
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

02 Oct 2018, 20:35
1
PriyankaGehlawat wrote:
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
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 non-negative 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 $$x=\sqrt{9}=3$$.
_________________
Intern
Joined: 16 Apr 2019
Posts: 9
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

### Show Tags

04 Jun 2019, 10:05
Dear Brunel,

You said-

"Is (x−3)2−−−−−−−√=3−x(x−3)2=3−x?

Remember: x2−−√=|x|x2=|x|. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0.

So x2−−√≥0x2≥0. But what does x2−−√x2 equal to?

Let's consider following examples:
If x=5x=5 --> x2−−√=25−−√=5=x=positivex2=25=5=x=positive;
If x=−5x=−5 --> x2−−√=25−−√=5=−x=positivex2=25=5=−x=positive."

My doubt is as follows-

All we know that sqrt of a number can be positive or negative results both, how you are saying "square root function can not give negative result"? If you kindly answer this question it would be a great help for me. Looking forward to hear you from.
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0   [#permalink] 04 Jun 2019, 10:05

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by