The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.

Let the two consecutive positive even numbers be x and (x + 2)

According to given condition,

x^{2} + (x + 2)^{2} = 452

x^{2} + x^{2} + 4x + 4 = 452 using (a + b)^{2} = a^{2} + 2ab + b^{2}

2x^{2} + 4x – 448 = 0

x^{2} + 2x – 224 = 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 = 2 c = – 224

= 1. – 224 = – 224

And either of their sum or difference = b

= 2

Thus the two terms are 16 and – 14

Difference = 16 – 14 = 2

Product = 16. – 14 = – 224

x^{2} + 2x – 224 = 0

x^{2} + 16x – 14x – 224 = 0

x(x + 16) – 14(x + 16) = 0

(x + 16) (x – 14) = 0

(x + 16) = 0 or (x – 14) = 0

x = – 16 or x = 14

x = 14 (x is positive odd number)

x + 2 = 14 + 2 = 16

Thus the two consecutive positive even numbers are 14 and 16

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