Q. 95.0( 1 Vote )

Find two consecutive multiples of 3 whose product is 648.

Answer :

Let the required consecutive multiples of 3 be 3x and 3(x + 1)

According to given condition,


3x.3(x + 1) = 648


9(x2 + x) = 648


x2 + x = 72


x2 + x – 72 = 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 = – 72


= 1. – 72 = – 72


And either of their sum or difference = b


= 1


Thus the two terms are 9 and – 8


Difference = 9 – 8 = 1


Product = 9. – 8 = – 72


x2 + 9x – 8x – 72 = 0


x (x + 9) – 8(x + 9) = 0


(x + 9) (x – 8) = 0


(x + 9) = 0 or (x – 8) = 0


x = – 9 or x = 8


x = 8 (rejecting the negative values)


3x = 3.8 = 24


3(x + 1) = 3(8 + 9) = 3.9 = 27


Hence, the required numbers are 24 and 27


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