The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.

Let the required consecutive multiples of 7 be 7x and 7(x + 1)

According to given condition,

(7x)^{2} + [7(x + 1)]^{2} = 1225

49 x^{2} + 49(x^{2} + 2x + 1) = 1225 using (a + b)^{2} = a^{2} + 2ab + b^{2}

49 x^{2} + 49x^{2} + 98x + 49 = 1225

98x^{2} + 98x–1176 = 0

x^{2} + x – 12 = 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 = 1 c = – 12

= 1. – 12 = – 12

And either of their sum or difference = b

= 1

Thus the two terms are 4 and – 3

Difference = 4 – 3 = 1

Product = 4. – 3 = – 12

x^{2} + 4x – 3x – 12 = 0

x(x + 4) – 3(x + 4) = 0

(x – 3) (x + 4) = 0

(x – 3) = 0 or (x + 4) = 0

x = 3 or x = – 4

when x = 3,

7x = 7.3 = 21

7(x + 1) = 7(3 + 1) = 7.4 = 28

Hence required multiples are 21, 28.

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