Puzzle #126: The Tricky Square Root

Here is another nice puzzle I saw on brilliant.org

Puzzle #126 Graphic

Puzzle #126 Graphic

The figure states it all…Happy rooting!

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6 Responses to Puzzle #126: The Tricky Square Root

  1. 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)

  2. Suman Saraf says:

    4 and -3

  3. Anant Aggarwal says:

    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)

  4. P R KUMAR says:

    Hi,

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

    Thanks & Regards,

    P R KUMAR

  5. Batman says:

    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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