Question

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Answer 1

Let the larger and smaller parts be x and y, respectively.
According to the question : 
`x+y=16`                             ...........(1) 

`2x^2=y^2+164`                ..............(2) 

From (i), we get: 

`x=16-y`                              ............(3) 

From (ii) and (iii), we get: 

`2(16-y)^2=y^2+164` 

⇒`2(256-32y+y^2)=y^2+164` 

⇒`512-64y+2y^2=y^2+164` 

⇒`y^2-64y+348=0` 

⇒`y^2-(58+6)y+348=0` 

⇒`y^2-58y-6y+348=0` 

⇒`y(y-58)(y-6)=0` 

⇒`y-58=0  or  y-6=0` 

⇒`y=6 (∵y<16)` 

Putting the value of y in equation (3), we get 

`x=16-6=10` 

Hence, the two natural numbers are 6 and 10.