## Puzzle #126: The Tricky Square Root

Here is another nice puzzle I saw on brilliant.org

The figure states it all…Happy rooting!

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### 6 Responses to Puzzle #126: The Tricky Square Root

1. Pratik Poddar (@pratikpoddar) says:

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)

• Alok Goyal says:

Great!

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.