Answer :

Let the smaller part be x and the larger part be 16 – x.

Given:

2 × (Larger Part)^{2} = (Smaller Part)^{2} + 164

2 × (16 – x)^{2} = (x)^{2} + 164

2 × (256 – 32x + x^{2}) = x^{2} + 164

512 – 64x + 2 x^{2} = x^{2} + 164

x^{2} – 64x + 512 – 164 = 0

x^{2} – 64x + 348 = 0

On factorizing the above equation,

Sum = -64

Product = 348

Therefore the two numbers satisfying the above conditions are -58 and -6.

x^{2} – 6x - 58x + 348 = 0

x(x – 6) - 58(x – 6) = 0

(x – 6) (x - 58) = 0

Solving first part,

x – 6 = 0

a = 6

Solving second part,

x - 58 = 0

x = 58

Since x is the smaller it cannot be greater than 16. Hence x cannot be 58

So the smaller part is x = 6

So the larger Part = 16 – 6

= 10

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