If X is not equal to 0. Then Sqrt(x^2)/x is = ?

[Saurabh Malpani]

Source Gmat Prep

If X is not equal to 0. Then Sqrt(x^2)/x is = ?

a) -1
b) 0
c) 1
d) x
e) IxI / x ---Read as Mod(x)/x

My answer is C i.e X is equal to 1.

Reason:
Case 1: Let X be equal to 2

sqrt (2^2) /2 = 1

Case 2: Let X be equal to -2

Sqrt (-2)^2/ (-2) = -2/-2 =1

Can some please suggest the flaw in my reasoning and suggest the correct Method!!

[dbernst]

Saurabh,

The flaw in your reasoning is with negative values for x. You wrote

Case 2: Let X be equal to -2

Sqrt (-2)^2/ (-2) = -2/-2 =1


However, (-2)^2 = POS 4. Sqrt 4 = POS 2. 2/-2 = -1.

Since you have already proved b, c, and d incorrect, the correct answer is E.

[Saurabh Malpani]

Dan,

Sorry but I don't undestand your reasoning.....what I have learnt is that when we take Square root of a Number X ...it can result in a +ve or -ve value.


i.e Say how do you interpret Sqrt(36) it can be either -6 or +6 isn't it?

Simillarly Sqrt of X^2 will result in either -x or +x isn't it?
Please reason this out as if I am missing something!!

Thanks
Saurabh Malpani

Saurabh,

The flaw in your reasoning is with negative values for x. You wrote

Case 2: Let X be equal to -2

Sqrt (-2)^2/ (-2) = -2/-2 =1


However, (-2)^2 = POS 4. Sqrt 4 = POS 2. 2/-2 = -1.

Since you have already proved b, c, and d incorrect, the correct answer is E.

[dbernst]

Saurabh,

On the GMAT you are only concerned with the positive root of a number. It a strange distinction, but an important one. For example, on the GMAT, if x^2 = 36, then x = +/- 6. However, If x = Sqrt 36, then x = 6 (only the positive root). I hope this helps!
-dan

[Saurabh Malpani]

Dan,

Thank you very much It was a great help!! ---I was missing a concept and my exam is just round teh corner so it the help came at the right time thanks!

I checked OG as well and I think I will also like to add a point, which may be some help to other people.

OG 10 Etd. DS 260,

Is Sqrt [(x-5)^2]= 5-x

a) -x*mod(x) >0
b) 5-x>0

Solution given in OG.

Note that Srt (y^2) = mod(y) for every y. Since Sqrt [(x-5)^2] = mod(x-5)
the equivalent equation is mod(x-5) = 5-x

which is only true when 5-x>=0 i.e 5>= x.

So A and B are self sufficient hence answer is D.

Saurabh Malpani

Saurabh,

On the GMAT you are only concerned with the positive root of a number. It a strange distinction, but an important one. For example, on the GMAT, if x^2 = 36, then x = +/- 6. However, If x = Sqrt 36, then x = 6 (only the positive root). I hope this helps!
-dan

[GMAT 5/18]

Saurabh,

Just have a quick question/comment regarding the OG question you just posted:

Is Sqrt [(x-5)^2]= 5-x

I tried to rephrase this, and came up with x^2 - 10x +25 = x^2 - 10x + 25.

So, I came to the conclusion that [/u]Sqrt [(x-5)^2]= 5-x is true, and therefore the answer is YES, FOR ALL VALUEs OF X, Sqrt [(x-5)^2]= 5-x.

I tested this using a couple of integers:

Using the integer 1, (1-5)^2 = 16, and root16 = 4. This equals 5-1 =4.
Using the integer -5, (-5-5)^2 = 100, and root100 = 10. This equals 5--5 = 10.

So, I guess what I am saying is, do we even need Statements I and II to prove sufficiency? Based on my ramblings above, I think not. If anyone has input, please share.

Thanks!

[Saurabh Malpani]

Hi GMAT 5/18,

My the fallacy I found in your apporach is that by Squaring both the side you are only considering mod(5-x) i.e you are assuming that 5-x and x-5 are same.

Ok let's take an example of x=6 rt?

so LHS (Left hand side) becomes Sqrt(6-5)==> Sqrt(1) ==> 1

RHS (Right hand Side) is 5-6 =-1

Now in values you considered in your apporach were between 0 & 5 hence you were able to justify that we don't need statement 1 and 2. but what about the values that don't fall in the range from 0 to 5.

Did I make sense or did I confused you further???

Let me know how can I help you further!!

Thanks
Saurabh Malpani

Saurabh,

Just have a quick question/comment regarding the OG question you just posted:

Is Sqrt [(x-5)^2]= 5-x

I tried to rephrase this, and came up with x^2 - 10x +25 = x^2 - 10x + 25.

So, I came to the conclusion that [/u]Sqrt [(x-5)^2]= 5-x is true, and therefore the answer is YES, FOR ALL VALUEs OF X, Sqrt [(x-5)^2]= 5-x.

I tested this using a couple of integers:

Using the integer 1, (1-5)^2 = 16, and root16 = 4. This equals 5-1 =4.
Using the integer -5, (-5-5)^2 = 100, and root100 = 10. This equals 5--5 = 10.

So, I guess what I am saying is, do we even need Statements I and II to prove sufficiency? Based on my ramblings above, I think not. If anyone has input, please share.

Thanks!

[Saurabh Malpani]

Sorry it should not be "My Fallacy" (what I wrote in my previous post) but The fallacy I found*

Secondly ..just to add all the values you considered were 5-x>=0 i.e 5>= x.


Saurabh Malpani

Hi GMAT 5/18,

My the fallacy I found in your apporach is that by Squaring both the side you are only considering mod(5-x) i.e you are assuming that 5-x and x-5 are same.

Ok let's take an example of x=6 rt?

so LHS (Left hand side) becomes Sqrt(6-5)==> Sqrt(1) ==> 1

RHS (Right hand Side) is 5-6 =-1

Now in values you considered in your apporach were between 0 & 5 hence you were able to justify that we don't need statement 1 and 2. but what about the values that don't fall in the range from 0 to 5.

Did I make sense or did I confused you further???

Let me know how can I help you further!!

Thanks
Saurabh Malpani

Saurabh,

Just have a quick question/comment regarding the OG question you just posted:

Is Sqrt [(x-5)^2]= 5-x

I tried to rephrase this, and came up with x^2 - 10x +25 = x^2 - 10x + 25.

So, I came to the conclusion that [/u]Sqrt [(x-5)^2]= 5-x is true, and therefore the answer is YES, FOR ALL VALUEs OF X, Sqrt [(x-5)^2]= 5-x.

I tested this using a couple of integers:

Using the integer 1, (1-5)^2 = 16, and root16 = 4. This equals 5-1 =4.
Using the integer -5, (-5-5)^2 = 100, and root100 = 10. This equals 5--5 = 10.

So, I guess what I am saying is, do we even need Statements I and II to prove sufficiency? Based on my ramblings above, I think not. If anyone has input, please share.

Thanks!

[GMAT 5/18]

Yes, great explanation Saurabh, thanks! I see it now!