Q. 164.5( 2 Votes )

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

Answer :

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.


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