Question

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

Answer 1

Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition
2x² – (16 – x)² = 164
⇒ 2x² – (256 – 32x + x²) = 164
⇒ 2x² – 256 + 32x – x² = 164
⇒ x² + 32x – 256 – 164 = 0
⇒ x² + 32x – 420 = 0
⇒ x² + 42x – 10x – 420 = 0
⇒ x(x + 42) – 10(x + 42) = 0
⇒ (x + 42)(x – 10) = 0
Either x + 42 = 0,
then x = –42,
but it is not possible.
or
x - 10 = 0,
then x = 10
∴ Larger part = 10
and smaller part = 16 - 10 = 6.