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 Twitter account. ( Log Out / Change )
You are commenting using your Facebook account. ( Log Out / Change )
You are commenting using your Google+ account. ( Log Out / Change )
Connecting to %s
Notify me of new comments via email.
Enter your email address to follow this blog and receive notifications of new posts by email.
Join 1,608 other followers
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.