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# Is root{(x-3)^2}=3-x?

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16 Jan 2013, 04:04
1
vaivish1723 wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

$$\sqrt{(x-3)^2} = 3-x$$
$$|x-3| = 3-x$$
$$3-x >= 0$$
$$3>=x$$ x is less than or equal to 3
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Re: [square_root](X-3)^2[/square_root][/m] = 3 -X ?  [#permalink]

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06 Mar 2013, 21:19
2
sujit2k7 wrote:
This is a DS question ..

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

1) X # 3
2) -X|X| > 0

Only thing which the question is asking is 3-X positive
As Sqrt (X-3)^2
= X-3 if X-3 is positive
= 3-X if 3-X is positive

STAT1
is INSUFFICIENT as X# 3 doesn't tell anything about whether 3-X is positive or not.

STAT2
-X|X| > 0
since |X| is positive so
-X > 0
=> X <0
and if X< 0 then 3-X will be positive.
Hope it helps!
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Re: [square_root](X-3)^2[/square_root][/m] = 3 -X ?  [#permalink]

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06 Mar 2013, 21:43
2
sujit2k7 wrote:
This is a DS question ..

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

1) X # 3
2) -X|X| > 0

We know that $$\sqrt{X^2}$$ = |X|. Thus, the question stem is asking whether |X-3| = 3-X. This is possible only if (X-3) is negative or X<3.

From F.S 1, we have x is not equal to 3. Clearly Insufficient.

From F.S 2, we have -X|X|>0. Thus, as |X| is always positive, X has to be negative. Thus, if X is negative, it will always be less than 3. Sufficient.

B.
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18 Apr 2013, 16:04
vaivish1723 wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

Here is how i solved it:

1) x [/s]=[/s] 3. Not sufficient
2) -x|x|>0 => x<0 only then the given inequality holds

so [/square_root](s-3)^2[/square_root] => -(x-3) (as we know [/square_root]x^2[/square_root] = - (x) if x<0) and 3-x = -(x-3) so sufficient.

IMO B.
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24 Dec 2014, 02:13
vaivish1723 wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

Hi Bunuel,

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Should the part in red not just be "<"?
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24 Dec 2014, 07:24
dmgmat2014 wrote:
vaivish1723 wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

Hi Bunuel,

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Should the part in red not just be "<"?

No, because x=3 also satisfies $$\sqrt{(x-3)^2}=3-x$$.
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07 Jan 2016, 01:29
HI understand the square root concept.

But I ended up squaring both sides of equation and got the question as |x-3|=|3-x|.

Can someone explain why cant we square both the sides of equation to eliminate square root on lefthand side?
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10 Jan 2016, 07:01
seemachandran wrote:
HI understand the square root concept.

But I ended up squaring both sides of equation and got the question as |x-3|=|3-x|.

Can someone explain why cant we square both the sides of equation to eliminate square root on lefthand side?

Hope it helps.
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10 Jan 2016, 14:03
Thanks a lot Bunuel!!
I got the concept, i forgot to apply the concept i.e not to square when not sure about the sign.
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11 Jan 2016, 19:22
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is root{(x-3)^2}=3-x?

(1) x≠3
(2) −x|x|>0

When you modify the original condition and the question, it becomes n-th power root (A^n)=|A| when n=even, and |A|=A when A>=0, |A|=-A when A<0. So, |x-3|=3-x=-(x-3)? becomes x-3<0?, x<3?. There is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), x=/3-> x=2 yes, x=4 no, which is not sufficient.
For 2), -x|x|>0 -> x<0<3, which is yes and sufficient. Therefore, the answer is B.

 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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11 Feb 2017, 00:59
Let's paraphrase the question.
$$\sqrt{(x-3)^{2}}$$ = (3-x) ? is equal to is x-3 < 0 ?

Apparently,stmt (1) is insufficient.

Consider stmt (2) : -x|x| > 0
This inequality can be deduced to -x > 0.
Hence x < 0;we now know that x-3 <0.
Thus,it is sufficient.

Ans B
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25 Apr 2017, 09:19
VeritasPrepKarishma wrote:
mrcrescentfresh wrote:
I am having a hard time grasping why we cannot simplify the problem as:

((X-3)^2)^1/2 = (3 - X) to (X - 3)^2 = (3 - X)^2?

I know I have worked problems before where we have been able to solve it by squaring both sides, but when done in this scenario the answer is entirely different than the OA.

That is because if the question says:
Is 5 = -5?
And you do not know but you square both sides and get 25 = 25
Can you say then that 5 = -5? No!

I understand that you would have successfully used the technique of squaring both sides before but that would be in conditions like these:
Given equation: $$\sqrt{X} = 3$$
Squaring both sides: X = 9
Here you already know that the equation holds so you can square it. It will still hold. This is like saying:
It is given that 5 = 5.
Squaring both sides, 25 = 25 which is true.

This question is similar to the first case. It is asked whether $$\sqrt{((X-3)^2)} = (3 - X)$$?

LHS is positive because $$\sqrt{((X-3)^2)} = |X-3|$$
and by definition of mod, we know that
|X| = X if X is positive or zero and -X if X is negative (or zero).
Since |X-3| = - (X - 3), we can say that X - 3 <= 0 or that X <= 3
So the question is: Is X <= 3?
Stmnt 1 not sufficient.
But stmnt 2 says -X|X| > 0
This means -X|X| is positive.
Since |X| will be positive, X must be negative to get rid of the extra negative sign in front. So this statement tells us that X < 0. Then X must be definitely less than 3. Sufficient.

How did you get this?
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26 Apr 2017, 00:56
1
hannahkagalwala wrote:
VeritasPrepKarishma wrote:
mrcrescentfresh wrote:
I am having a hard time grasping why we cannot simplify the problem as:

((X-3)^2)^1/2 = (3 - X) to (X - 3)^2 = (3 - X)^2?

I know I have worked problems before where we have been able to solve it by squaring both sides, but when done in this scenario the answer is entirely different than the OA.

That is because if the question says:
Is 5 = -5?
And you do not know but you square both sides and get 25 = 25
Can you say then that 5 = -5? No!

I understand that you would have successfully used the technique of squaring both sides before but that would be in conditions like these:
Given equation: $$\sqrt{X} = 3$$
Squaring both sides: X = 9
Here you already know that the equation holds so you can square it. It will still hold. This is like saying:
It is given that 5 = 5.
Squaring both sides, 25 = 25 which is true.

This question is similar to the first case. It is asked whether $$\sqrt{((X-3)^2)} = (3 - X)$$?

LHS is positive because $$\sqrt{((X-3)^2)} = |X-3|$$
and by definition of mod, we know that
|X| = X if X is positive or zero and -X if X is negative (or zero).
Since |X-3| = - (X - 3), we can say that X - 3 <= 0 or that X <= 3
So the question is: Is X <= 3?
Stmnt 1 not sufficient.
But stmnt 2 says -X|X| > 0
This means -X|X| is positive.
Since |X| will be positive, X must be negative to get rid of the extra negative sign in front. So this statement tells us that X < 0. Then X must be definitely less than 3. Sufficient.

How did you get this?

No, we are not establishing/using this. We are just simplifying the question. Since the question is:

Is $$|X-3| = - (X - 3)$$ ?
Is $$(X - 3) <= 0$$?
Is $$X <= 3$$?

And then we go on to evaluate each statement.
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27 Apr 2017, 23:39
vaivish1723 wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

quite simple , as it is known that root could not have -ve values so we are sure about the LHS

now go through the statements-
statement 1. x is not equal to 3 so x can be anything except 3 , try plugging in say x=5 then LHS(2) not equal to RHS(-2) so we get A No

say x=-5 then LHS(8) = RHS(8) so we get A yes . so not sufficient

statement 2. it says that x must be -ve so we get A yes so sufficient

hence B
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05 May 2017, 00:55
(x-3)^2}=3-x==>> means should be always positive

is 3-x>0 or x<3 ?

(1) x≠3 ,so x can>3 as well so INSUFF

(2) −x|x|>0-->

TO BE GREATER THAN 0 X SHOULD BE -VE ALWAYS
Hence SUFF
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30 Aug 2017, 07:01
Bunuel
niks18 amanvermagmat

Quote:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

With this understanding I proceeded a bit differently and want to know if my approach is correct:

If I look closely at part inside sq root in LHS, then RHS = -(LHS)

With this thinking clearly i need term inside |x|to be negative since - (x-3) = 3 - x
since |x| =$$\sqrt{x^2}$$

Proceeding further for statement 2, -x * |x| is a positive value, this is possible only when
product is of two negative no
so essentially |x| = -x or in other words - (x-3) = 3-x which is our RHS
Does this makes sense?
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02 Feb 2018, 13:32
Since the question doesnt say anything about x being an integer only, Why arent we testing the fraction use cases which comply with the DS statements ?

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02 Feb 2018, 20:08
1
Bunuel
niks18 amanvermagmat

Quote:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

With this understanding I proceeded a bit differently and want to know if my approach is correct:

If I look closely at part inside sq root in LHS, then RHS = -(LHS)

With this thinking clearly i need term inside |x|to be negative since - (x-3) = 3 - x
since |x| =$$\sqrt{x^2}$$

Proceeding further for statement 2, -x * |x| is a positive value, this is possible only when
product is of two negative no

so essentially |x| = -x or in other words - (x-3) = 3-x which is our RHS
Does this makes sense?

i am a bit concerned about the above highlighted statement. product of two numbers can be positive when both are positive.
we know |x| is always positive so for -x|x| to be positive -x>0 =>x<0 i.e x has to be negative

now if LHS $$\sqrt{(x-3)^2}$$. we know that square root is always positive so LHS is positive

RHS=3-x where x is negative i.e. -x is positive so RHS=positive+positive=positive. Hence LHS=RHS

overall seems to be ok
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02 Feb 2018, 20:10
akashyap2202 wrote:
Since the question doesnt say anything about x being an integer only, Why arent we testing the fraction use cases which comply with the DS statements ?

Hi akashyap2202

In this question we simply need to know whether x<0. Now x can be an integer or a fraction. testing for fraction will not make any difference.
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21 Jul 2019, 12:35
Bunuel wrote:
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.

Hi Bunuel - I solved the Q by the following approach and got the anser as option D. Can you please help me understand where am I going wrong.
Q stem - LHS - (x-3)^2*1/2 = (x-3). RHS = (3-x) so LHS = RHS only when x=3. St 1 clearly tells us that x is not equal to 3 so it is sufficient.
St 2 is -x|x| >0. Now multiplying both sides by -1 will give x|x|< 0 which means that x has to be negative since |x| can not be negative. Going back to the Q stem, this would mean that LHS (as simplified earlier) - (x-3) will be negative and (3 - x) will become (3 - (-x)) which will be positive so LHS is not eqaul to RHS, therefore st 2 is sufficient
Re: Is root{(x-3)^2}=3-x?   [#permalink] 21 Jul 2019, 12:35

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