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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 x2 + 49(x2 + 2x + 1) = 1225 using (a + b)2 = a2 + 2ab + b2
49 x2 + 49x2 + 98x + 49 = 1225
98x2 + 98x–1176 = 0
x2 + 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
x2 + 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.