# The sum of two na

Let the required number be x and 28 – x

According to given condition,

x(28 – x) = 192

x2 – 28x + 192 = 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 = – 28 c = 192

= 1.192 = 192

And either of their sum or difference = b

= – 28

Thus the two terms are – 16 and – 12

Sum = – 16 – 12 = – 28

Product = – 16. – 12 = 192

x2 – 28x + 192 = 0

x2 – 16x – 12x + 192 = 0

x(x – 16) – 12(x – 16) = 0

(x – 16) (x – 12) = 0

(x – 16) = 0 or (x – 12) = 0

x = 16 or x = 12

Hence the required numbers are 16, 12

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