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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

I get it now. Actually, I didn't at first, then stumbled upon another question, a geometry one this time, and could draw the parallels. The question here doesn't ask if x<=0, it asks if x<3.

Got it, thanks. The geometry question I am referring to is this - a-circle-is-drawn-on-a-coordinate-plane-if-a-line-is-drawn-161692.html (the question asks if slope is less than 1, not zero, not anything else)!

Thanks for the great explanations and the GMAT Club Math Book as well! Great resource.

A question regarding the above question and square roots/absolute values in general. In the GMAT Club Math Book you write "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."

What you are basically saying here is that whenever we have an equation with a positive and even square root we shall utilize the absolute value. BUT when we have just a random figure, we shall use only the positive root. Right?

Thanks again! _________________

Thank you very much for reading this post till the end! Kudos?

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,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help

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,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help

The question asks whether x\leq{3}? The second statement says that x<0. So, the answer to the question is yes.

Also, if we know that x < 0, then HOW can x be 2? _________________

Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]
23 Oct 2014, 22:40

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. ?

Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]
24 Oct 2014, 02:21

1

This post received KUDOS

Expert's post

arpitsharms wrote:

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.

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