Here is another nice puzzle I saw on brilliant.org

Puzzle #126 Graphic

The figure states it all…Happy rooting!

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)

4 and -3

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)

Great!

Hi,

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

Thanks & Regards,

P R KUMAR

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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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)

4 and -3

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)

Great!

Hi,

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

Thanks & Regards,

P R KUMAR

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.