# A two – digit num

Let the digits at units and tens place be x and y respectively

xy = 14 – – – – (1)

According to the question

(10y + x) + 45 = 10x + y

9y – 9x = – 45

y – x = – 5 – – – – – (2)

From (1) and (2) we get  14 – x2 = – 5x

x2 – 5x – 14 = 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 = – 5 c = – 14

= 1. – 14 = – 14

And either of their sum or difference = b

= – 5

Thus the two terms are – 7 and 2

Difference = – 7 + 2 = – 5

Product = – 7.2 = – 14

x2 – 5x – 14 = 0

x2 – 7x + 2x – 14 = 0

x(x – 7) + 2(x – 7) = 0

(x + 2)(x – 7) = 0

x = 7 or x = – 2

x = 7 (neglecting the negative part)

Putting x = 7 in equation (1) we get

y = 2

Required number = 10.2 + 7 = 27

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