Question

Divide 16 into two parts such that 2 times the square of larger part is 164 more than the square of smaller part ?

Answer 1

Hey mate!!!
Here is youranswer.
Let larger part be=x
Let smaller part be=y
x+y=16
y=16-x (1)
2(x)^2=164+(16-x)^2 (using 1)
2x^2=164+256+x^2-32x
2x^2=420+x^2-32x
2x^2-x^2+32x-420=0
x^2+32x-420=0
x^2+42x-10x-420=0
x(x+42)-10(x+42)=0
(x-10)(x+42)=0
x=10, x=-42
We will reject -42.
Larger part =x=10 and
smaller part =y=16-x=16-10=6.
Hope it helps!!!!!!

Answer 2

Let one of the parts be x and the other part be 16 - x
Also, let the larger part be x.
Given that two times the square of the larger part is 164 more than the square of the smaller part.
So,
2x² = (16 - x)² + 164
2x² = 256 + x² - 32x + 164
2x² - x² + 32x = 420
x² + 32x - 420 = 0
x² + 42x - 10 - 420 = 0
x(x + 42) - 10(x + 42) = 0
(x + 42) (x - 10) = 0
x = - 42 or x = 10.

Rejecting x = - 42 as it did not satisfy the two parts of 16. 
So,
x = 10
the other part is 16 - 10 = 6

The two parts are 10 and 6.

Hope this helps you.