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

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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 22 Nov 2014, 07:20
1
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

Am I right In my logic.Please help


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.

Answer: B.

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.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 22 Nov 2014, 08:26
okay gr8 ...
I have a doubt about this we have sqrt of x is always greater than equal to 0 .



Now same concept is what we applied on the above problem.

But there is one problem that u explained on a GMAT PREP question .

Let's consider following examples:

I am concern about the second one when x=-5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=-x
I tried posting this on the


BUT was not accepting the expression ...:( havent yet learnt to use expression ...


thanks in advance.


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

Am I right In my logic.Please help


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.

Answer: B.

Hope it helps.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 22 Nov 2014, 09:09
hanschris5 wrote:
okay gr8 ...
I have a doubt about this we have sqrt of x is always greater than equal to 0 .



Now same concept is what we applied on the above problem.

But there is one problem that u explained on a GMAT PREP question .

Let's consider following examples:

I am concern about the second one when x=-5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=-x
I tried posting this on the


BUT was not accepting the expression ...:( havent yet learnt to use expression ...


thanks in advance.


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

Am I right In my logic.Please help


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.

Answer: B.

Hope it helps.


\(\sqrt{x^2}=\sqrt{25}=5\) --> x = 5 or x = -5. The square root gives positive result but x itself can be 5 or -5. What is confusing there?
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 22 Nov 2014, 10:26
okay so its only do with the result ie the x that comes out as result after the operation ... whatever it may have been inside the expression...

thanks. doubt cleared.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 27 Nov 2015, 09:11
Given to us \(\sqrt{(x-3)^2} = 3-x\)

As a general principle, whenever we are given square root of a square of a number, we can have 2 solutions for such a number. It is illustrated below:

\(\sqrt{x^2} = x (if x>=0)\)
\(\sqrt{x^2} = -x (if x<=0)\)
This square root of a square of a number is essentially the modulus of number and thus always gives a positive answer.

So in questions like the one illustrating the above principle we have to consider both the possibilities.

Now, for the given equation:

The two possibilities are

1st possibility : \(\sqrt{(x-3)^2} = x-3\) (if x>=3)
2nd possibility : \(\sqrt{(x-3)^2} = 3-x\) (if x<=3)

Now considering the given statements:

Using Statement 1

if \(x\neq{3}\)

This statement satisfies both the possibilities
Hence we cannot reach to a single possibility, so statement 1 is not sufficient.

Using Statement 2

\(-x|x| > 0\)

For this condition to be true, x needs to be necessarily less than 0.

Now if x<0, then it definitely is less than 3, hence will satisy the 2nd possibility, thus, giving a distcict answer.

Hence statement 2 is sufficent to answer \(\sqrt{(x-3)^2} = 3-x\)

So the answer is B.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 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.

The correct answer is B.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 02 Feb 2017, 02:51
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

Am I right In my logic.Please help


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.

Answer: B.

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.

Whatever you put into the |x|, it always gives you positive value. If you multiply 2 positive or 2 negative values, it always gives you positive result, which is ''some expression''>0. Here, |x| already gives you positive value. So, the green part has to be negative to legit the statement. Here, is test case:
-x*positive value>0
--> -(-10)*positive value>0
-->positive value*positive value>0
So, to legit the statement 2, x must be negative. The statement is sufficient.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 23 Apr 2017, 20:01
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

Am I right In my logic.Please help


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.

Answer: B.

Hope it helps.


Hi Bunuel,

From \(|x-3|=3-x\), since we know |x-3| is positive, can we not go one step further and simplify the equation to:

x-3=3-x
=>x=3?

What am I doing wrong?
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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New post 23 Apr 2017, 22:44
puto 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

Am I right In my logic.Please help


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.

Answer: B.

Hope it helps.


Hi Bunuel,

From \(|x-3|=3-x\), since we know |x-3| is positive, can we not go one step further and simplify the equation to:

x-3=3-x
=>x=3?

What am I doing wrong?


I tried to explain exactly this point in may solutions. From |x - 3| = 3 - x, when x - 3 <= 0. If it's still unclear you should go through theory about absolute values:

Theory on Absolute Values: http://gmatclub.com/forum/math-absolute ... 86462.html
Absolute value tips: http://gmatclub.com/forum/absolute-valu ... 75002.html

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]

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Re: Is sqrt ((x-3)^2) = 3-x?   [#permalink] 25 Apr 2018, 06:00

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