Q. 203.9( 13 Votes )

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Answer :

Let the larger and the smaller parts be x and y respectively.

According to the question


x + y = 16 – – – – – (1)


2x2 = y2 + 164 – – – (2)


From (1) x = 16 – y – – – (3)


From (2) and (3) we get


2(16 – y)2 = y2 + 164


2(256 – 32y + y2) = y2 + 164 using (a + b)2 = a2 + 2ab + b2


512 – 64y + 2y2 = y2 + 164


y2 – 64y + 348 = 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 = – 64 c = 348


= 1. 348 = 348


And either of their sum or difference = b


= – 64


Thus the two terms are – 58 and – 6


Sum = – 58 – 6 = – 64


Product = – 58. – 6 = 348


y2 – 64y + 348 = 0


y2 – 58y – 6y + 348 = 0


y(y – 58) – 6(y – 58) = 0


(y – 58) (y – 6) = 0


(y – 58) = 0 or (y – 6) = 0


y = 6 (y < 16)


putting the value of y in (3), we get


x = 16 – 6


= 10


Hence the two natural numbers are 6 and 10.


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