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

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

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

I think there's a slight flaw in the solution. Is sqrt((x-3)^2) = |x-3|? Arent we eliminating the negative square root here?

I think the answer is D. Substitute -5 as an example, LHS is + -8 and RHS is +8

(or am I missing something here?)

First of all \(\sqrt{x^2}\) does equal to \(|x|\), so \(\sqrt{(x-3)^2}=|x-3|\).

Next GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

And finally (1) can not be sufficient as if you try \(x=5\neq{3}\), then \(LHS=\sqrt{(x-3)^2}=\sqrt{(5-3)^2}=2\neq{RHS}=3-x=3-5=-2\).

Hope it's clear.

For \({\sqrt{(X-3)^2} = 3-X\) to be true, \(3-X\) will always have to be positive or 0, because in GMAT a root can not be negetive. So \(3-X>0\) or \(X<3\).

S1:\(X =! 3\), but is it less than 3? we dont know. Not sufficient.
S2: \(-X|X|>0\), so either both (\(-X\) and \(|X|\)) are -ive or both are +ive. \(|X|\) can not be negetive so \(-X\) must be +ive, or \(X\) must be negetive, which would clearly be less than 3. Sufficient.

Answer: B

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.

Hi!

Statement (2) reads as "the product of (negative x) and (absolute value of x) is positive".

Since |x| is always non-negative (i.e. |x|=0 when x=0 and is positive for all other values of x), in order for statement (2) to be true, x must be negative, giving us:

-(-) * |-| > 0

+ * + > 0

which is true.

So, basically, (2) is a fancy way of saying:

x < 0.

Now that you know how to interpret (2), try the question again - if you still need a hand, let us know where you get stuck!

Warlock007: As per your request, I am explaining the solution to this question. The solution is exactly as given by walker above. I will just elaborate on 'why'

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

1)\(x\ne3\)

2) \(-x*|x| >0\)

When we say square root, we mean just the positive square root. Another way to think about it is:

\(\sqrt{x^2} = |x|\) i.e. square root is only the positive root.

Therefore, \(\sqrt{(x-3)^2} = |x - 3|\)
Now the question is, is |x - 3| = 3 - x?

Now the definition of modulus can be used here.
|x| = x when x >= 0
and |x| = -x when x < 0

i.e. |5| = 5 since 5 > 0
and |-5| = -(-5) = 5 since -5 < 0 (Works for x = 0 too)

Therefore, |x - 3| = -(x - 3) = (3 - x) if (x - 3) <= 0

Let's go back to our question: Is |x - 3| = 3 - x?
|x - 3| = 3 - x only when (x - 3) <= 0 i.e. when x <= 3

Stmnt 1: \(x\ne3\)
No idea whether x is less than 3 so not sufficient.

Stmnt 2: -x*|x| >0
Now, mods are always greater than or equal to 0 (i.e. they can never be negative)
So |x| has to be positive.
Then -x must be positive too to make -x*|x| >0
This means x must be negative (only then will -x be positive)
If x is negative, it is certainly less than 3. Hence, this stmnt alone is sufficient.

Answer (B)

You could also do this question by plugging in numbers and checking (say try x = -1, 0, 1, 3, 4) but plugging in for such questions makes me worry that I have forgotten to check for some specific condition and hence I avoid doing it.

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.

The answer is therefore (B).

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.
So, Answer will be B
Hope it helps!

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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.

Hi Bunuel
I am still confused about this.Please help me out.
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.

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.

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 \(x=\sqrt{9}=3\).

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.