Question

X is the any natural number is greater than 3 times the squreroot of it by 4 then find the number

Answer 1

Let the no. be x

According to question

x=3 root x + 4

x-4 = (3 root x^2)

x^2- 8x + 16 = 9x

x^2 - 8x-9x +16 = 0

x^2 -17x +16= 0

(x-16)(x-1) =0

x = 16, x = 1

x= 1 is not possible

So, answer is 16