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 : 
${\displaystyle x+y=16}$                             ...........(1) 

${\displaystyle 2x^2=y^2+164}$                ..............(2) 

From (i), we get: 

${\displaystyle x=16-y}$                              ............(3) 

From (ii) and (iii), we get: 

${\displaystyle 2{(16-y)}^2=y^2+164}$ 

⇒${\displaystyle 2(256-32y+y^2)=y^2+164}$ 

⇒${\displaystyle 512-64y+2y^2=y^2+164}$ 

⇒${\displaystyle y^2-64y+348=0}$ 

⇒${\displaystyle y^2-(58+6)y+348=0}$ 

⇒${\displaystyle y^2-58y-6y+348=0}$ 

⇒${\displaystyle y(y-58)(y-6)=0}$ 

⇒${\displaystyle y-58=0 \text{or} y-6=0}$ 

⇒${\displaystyle y=6(\because y<16)}$ 

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

${\displaystyle x=16-6=10}$ 

Hence, the two natural numbers are 6 and 10.