Question

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

Answer 1

Let the smaller number be x.
Larger number= 16-x
2(16-x)^2-x^2=164
2(256-32x+x^2)=164+x^2
512-64x+2x^2=164+x^2
512-64x+2x^2-164-x^2=0
x^2-64x+348=0
x^2-58x-6x+348=0
x(x-58)-6(x-58)=0
(x-6)(x-58)=0
Therefore, x can be 58 or 6.
Since, 16 can not be divided into 58
therefore,x=6

Answer 2

Let the smaller number be x.

Larger number= 16-x

2(16-x)^2-x^2=164

2(256-32x+x^2)=164+x^2

512-64x+2x^2=164+x^2

512-64x+2x^2-164-x^2=0

x^2-64x+348=0

x^2-58x-6x+348=0

x(x-58)-6(x-58)=0

(x-6)(x-58)=0

Therefore, x can be 58 or 6.

Since, 16 can not be divided into 58

therefore,x=6