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.

Fill in your details below or click an icon to log in:

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

Notify me of new comments via email.

Notify me of new posts via email.

Enter your email address to follow this blog and receive notifications of new posts by email.

Join 1,715 other followers

Follow

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.