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

Let the larger and the smaller parts be x and y respectively.

According to the question

x + y = 16 – – – – – (1)

2x^{2} = y^{2} + 164 – – – (2)

From (1) x = 16 – y – – – (3)

From (2) and (3) we get

2(16 – y)^{2} = y^{2} + 164

2(256 – 32y + y^{2}) = y^{2} + 164 using (a + b)^{2} = a^{2} + 2ab + b^{2}

512 – 64y + 2y^{2} = y^{2} + 164

y^{2} – 64y + 348 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = – 64 c = 348

= 1. 348 = 348

And either of their sum or difference = b

= – 64

Thus the two terms are – 58 and – 6

Sum = – 58 – 6 = – 64

Product = – 58. – 6 = 348

y^{2} – 64y + 348 = 0

y^{2} – 58y – 6y + 348 = 0

y(y – 58) – 6(y – 58) = 0

(y – 58) (y – 6) = 0

(y – 58) = 0 or (y – 6) = 0

y = 6 (y < 16)

putting the value of y in (3), we get

x = 16 – 6

= 10

Hence the two natural numbers are 6 and 10.

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