If r,s and t are integers, is r+s =square root of t

Question

If  r,s  and  t  are integers, is  r+s =\sqrt{t}  ?

r<-s

(r+s)^2=t

Tulkin987 · Accepted Answer

(1)  r<-s  
 r+s<0 . Since,  \sqrt{t}  is non-negative  positive , answer is NO. Sufficient.

(2)  |r+s|=\sqrt{t} . Not sufficient.

Answer: A

Princ · Answer

OA: A

Statement 1 :  r<-s 
 r+s<0 
 \sqrt{t}  can not be -ve. 
So answer for  is  r+s=\sqrt{t}   :no.
Statement 1 is alone sufficient.

Statement 2 :  (r+s)^2=t 
 |r+s|=\sqrt{t}

Case a :When  r+s <0 
Then statement 2 becomes  -(r+s)=\sqrt{t} 
Answer for  is  r+s=\sqrt{t}   :no

Case b : When  {r+s}\geq{0} , 
then statement 2 becomes  r+s=\sqrt{t} . 
answer for  is  r+s=\sqrt{t}   :yes.
So statement 2 alone is insufficient.

GMATBusters · Answer

(1)  r<-s  
 r+s<0 . So, r+s is negative. Since,  \sqrt{t}  is always  NON NEGATIVE , answer is NO. 
 Sufficient.

(2)  |r+s|=\sqrt{t} . 
Now  (r+s)  can be equal to  \sqrt{t}  or - \sqrt{t}. 
 Not sufficient.

Answer: A

GYANENDRA88 · Answer

gmatbusters 
LET t=36
ROOT36=+- 6
ROOT(t) can be either +ve or -ve

EgmatQuantExpert · Answer

Let me clarify it with two specific cases:

• If  a^2  = 16, then a = +4 or -4 
It happens because from the equation  a^2  = 16, we can derive (a + 4)(a – 4) = 0, which gives us two values of a = +4 and -4
This can also be written as a = + \sqrt{16}  or - \sqrt{16} 
Now,  \sqrt{16}  = 4, 
 Hence, the values of a = +4 or -4

• If a = 16, then  \sqrt{a}  =  \sqrt{16}  = 4 always. Therefore  \sqrt{a}  is always positive.

Hope this clarifies your doubt.

GYANENDRA88 · Answer

EgmatQuantExpert 
Thanks for your reply.
One can approach it in a different way-
let a=-6
 t=a^2
 root(t)=root(a^2)=-6

GMATBusters · Answer

Hii
I am afraid your understanding is not correct.
If x² = 36, x = +/-6
But if x =  \sqrt{36} 
x = only 6.
 The square root function gives only the positive value. 
This is as per the definition of sqaure root function.

Please remember this concept, this is very important for GMAT.

If you have any further queries, feel free to tag me.

Happy Learning...

GYANENDRA88 · Answer

gmatbusters 
Thanks for the reply
can you prove it mathematically that root36 is not equal to -6.

GMATBusters · Answer

Hii A function is a special relationship where each input has a single output. So to define square root function, root function must give a unique answer. it has been defined as to be positive. . There is no reason for this. See it like this, we say that the charge of electron is negative, actually it is defined to be negative. Scientists might have defined charge of electron to be positive and proton charge as negative. But who knows, God might be in favour of Proton that day.

Similarly we say that numbers on the number line to left of zero is negative and to right of zero is positive. We could have defined the convention otherwise also. But it is defined like this. There is no explanation for this. Hope it is clear now.

Princ · Answer

The symbol √ does not mean square root. It means "principal square root", which is the positive square root. From GMAT prespective , check out bunuel's comment Same matter has been discussed in below link https://gmatclub.com/forum/square-root-always-positive-114114.html You can also check below link from Manhattanprep :What’s the Deal with Square Roots on the GMAT? https://www.manhattanprep.com/gmat/blog ... -the-gmat/

GMATBusters · Answer

Hi  Tulkin987

We can not say that  \sqrt{t}  is positive, In fact  \sqrt{t}  is  NON NEGATIVE.

Tulkin987 · Answer

Hi, gmatbusters ! Corrected the mistake. Thanks

amoghhlgr · Answer

Bunuel  Lets say,  r + s = \sqrt{t}

Squaring on both sides,  (r+s)^{2} = (\sqrt{t})^{2} ... so  (r+s)^{2} = |t| ... now  (r+s)^{2} = t , since t cannot be negative. So, B alone looks sufficient. Can you please point at the mistake in this line of reasoning?

Thank you!

bumpbot · Answer

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If r,s and t are integers, is r+s =square root of t

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If r,s and t are integers, is r+s =square root of t

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