6 Responses to Puzzle #126: The Tricky Square Root

1. Pratik Poddar (@pratikpoddar) says:

   January 2, 2016 at 10:50 pm

   Ans = 4.

   Lets say that the value is x.
   sqrt (12+x) = x
   x^2 – x – 12 = 0
   (x-4)(x+3) = 0
   x = 4 (since x>0)

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2. Suman Saraf says:

   January 2, 2016 at 10:59 pm

   4 and -3

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3. Anant Aggarwal says:

   January 3, 2016 at 6:51 am

   I remember, 15 years ago, this was a very popular question in IIT JEE screening.

   (12+x)^(1/2) = x
   x^2 -x – 12 = 0
   x = 4 (cannot be the negative root of quadratic equation as that would imply (12)^(1/2) as an imaginary number)

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   • Alok Goyal says:

     January 3, 2016 at 8:10 am

     Great!

     Reply
4. P R KUMAR says:

   January 3, 2016 at 1:34 pm

   Hi,

   Please see the attached file for the solution to puzzle no. 126.

   Thanks & Regards,

   P R KUMAR

   Reply
5. Batman says:

   January 10, 2016 at 11:01 am

   Let the given expression be x. Squaring both sides we get x^2=x+12. This quadratic factorizes to (x-4)(x+3)=0. Yielding the possible values of x to be x= -3 and x=4. Since x is positive we clearly have x =4 as the only answer.

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