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# Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3

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Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink]

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20 Dec 2009, 07:50
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Is $$\sqrt{(x-3)^2}=3-x$$?

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

(2) -x|x| > 3
[Reveal] Spoiler: OA
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20 Dec 2009, 18:25
Ans: E.

First statement is not sufficient because if x=4, it may or may not be true. LHS = +1 or -1 while RHS = -1.

Second statement leads us to -3^1/2 < x < 0. For all the values between 0 and -3^1/2, which leads to |LHS| = RHS.
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21 Dec 2009, 02:47
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jan4dday wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in

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

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

When $$x>3$$, then RHS 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 is not equal to 3. Clearly insufficient.

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

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Re: How to solve this Question...I have no clue... [#permalink]

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27 Feb 2010, 21:47
is \sqrt{(x-3)^2} = 3-x?
square root of a number is positive.
if x<=3 .. 3-x is always positive
if x>3 ... 3-x is negative

st 1) x!=3 .
not sufficient as we dont know if x>3 or x<3
st 2) -x |x| > 0
|x| is always greater than 0, so x has to be negative for the expression to be > 0

if x is negative, then \sqrt{(x-3)^2} = 3-x

B
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Re: How to solve this Question...I have no clue... [#permalink]

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17 Mar 2011, 03:58
Just a slight correction I think...
St2 is saying x<-sqrt of 3 which is still a sufficient condition.
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17 Mar 2011, 04:27
I got x<-1.732 from statement 2

This satisfies the x<3 requisite
Hence B

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Re: How to solve this Question...I have no clue... [#permalink]

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17 Mar 2011, 23:07
sqrt((x-3)^2) = |x-3| = x-3 if x > 3 or 3-x if x < 3

(1) is not enough

(2) -x|x| is positive then

So -x is also +ve and hence x is -ve

so x < 3, hence answer is B
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Re: How to solve this Question...I have no clue... [#permalink]

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18 Mar 2011, 20:49
I'm a bit confused by this problem
Why can I not square both sides of the statement to get rid of the radical?
I will then have (x-3)^2=(3-x)^2
Statement 1 and 2 both will not apply in this case and the answer is then E.

However if I choose Bunuels method of doing it, I still dont get how B is the answer.
Statement 2 basically says that x=-2,-3....and so on and so forth will satisfy the statement. So basically all negative numbers equal to or less than 2 will satisfy the condition...
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Is (x - 3) ^2 = 3 - x 1) x not equal to 3 2) -x|x| [#permalink]

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24 Jul 2011, 22:55
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Is $$\sqrt{(x - 3)}^2$$ = 3 - x

1) x not equal to 3

2) -x|x| > 0
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Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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24 Jul 2011, 23:17
s = 3 - x

1) x not equal to 3

2) -x|x| > 0

Ans: Q can be repharsed as is 3-x>0 ?

Stmt1: x<>3 --> x can take on any values other that x such as x = -9 and 3-(-9) = 12

or x = 10 and 3-(10) = -7 Insuff

Stmt 2 : x has to be negative and any negative value of x will render 3 - x positive. Suff

(B)
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Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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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.

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Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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28 Jul 2011, 10:26
could you explain me please why do you use inequality signs? in question stem there is just equal sign
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Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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28 Jul 2011, 22:48
The inequality sign is needed to work with the modulus. When we remove the square root sign, we must take the modulus of the answer because the answer could be positive or negative.

To know more about working with modulus operators, please search for 'Absolute value (Modulus'
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Is square_root of (x-3)^2 = 3 - x ? [#permalink]

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27 Jun 2012, 12:08
Is $$\sqrt{(x-3)^2} = 3 - x$$ ?

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

According to my reasoning:
$$\sqrt{(x-3)^2}$$$$= |x-3|$$
So, the question is:
Is $$|x-3| = 3 -x$$?
In other words:
Is $$x - 3 < 0$$ ?
Is $$x < 3$$?

So, let's see the clues:
(1) x is not equal to 3
INSUFFICIENT.

(2) $$-x |x| > 0$$
Based on this info we know that x is not zero.
So,
$$-x > 0$$
$$x < 0$$
SUFFICIENT.

IMO, the OA is :
[Reveal] Spoiler:
B

Please, confirm whether I am Ok.
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Re: Is square_root of (x-3)^2 = 3 - x ? [#permalink]

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27 Jun 2012, 12:15
Merging similar topics.

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

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

According to my reasoning:
$$\sqrt{(x-3)^2}$$$$= |x-3|$$
So, the question is:
Is $$|x-3| = 3 -x$$?
In other words:
Is $$x - 3 < 0$$ ?
Is $$x < 3$$?

So, let's see the clues:
(1) x is not equal to 3
INSUFFICIENT.

(2) $$-x |x| > 0$$
Based on this info we know that x is not zero.
So,
$$-x > 0$$
$$x < 0$$
SUFFICIENT.

IMO, the OA is :
[Reveal] Spoiler:
B

Please, confirm whether I am Ok.

You are right OA is B.
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28 Jun 2012, 14:53
Bunuel wrote:
jan4dday wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in

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

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

When $$x>3$$, then RHS 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 is not equal to 3. Clearly insufficient.

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

Dear Bunuel,
is it possible in this question to solve by taking conditions x>0 and x<0 for |x-3| = 3-x instead of x>3 ?
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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink]

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21 Nov 2012, 22:02
arvindbhat1887 wrote:
I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help

The question is

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

(1) x<>3

(2) -x|x|>0

$$\sqrt{y}$$ can only be positive or 0.

So,The question is basically asking whether 3-x >= 0

1) Insufficient. If x=1, true, If x=10, false.

2) From this we get that x is negative. So -x is postivie. So, 3 + (-x) will always be a positive number. Sufficient.

Kudos Please... If my post helped.
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09 Jan 2013, 19:14
Bunuel wrote:
jan4dday wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in

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

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

When $$x>3$$, then RHS 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 is not equal to 3. Clearly insufficient.

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

(2) Actually $$-x|x| > 3$$, implies that $$x<-sqrt(3)$$, but still, as you mentioned, consequently implies that $$x<3$$. Sufficient.
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31 Mar 2014, 12:55
Hi Bunuel,

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

Regards,
Ravi
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31 Mar 2014, 13:22
email2vm wrote:
Hi Bunuel,

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

Regards,
Ravi

I got it now!

|x-3| = (3-x) ===> |3-x| = (3-x) ===>which is only possible when (3-x) >=0 (i.e +ve)

(like |x| =x only when x>=0)

so we have 3-x >=0 ====> x<=3

Also for option B

-x|x| > 3
==> x|x| < 3
so |x| is always positive but so make it less than 3 , x must be negative. i.e. x<0
Re: TOUGH DS Q!   [#permalink] 31 Mar 2014, 13:22

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