Q. 234.0( 6 Votes )

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.

Answer :

Let the three consecutive positive integers be x, x + 1, x + 2

According to the given condition,


x2 + (x + 1)(x + 2) = 46


x2 + x2 + 3x + 2 = 46


2x2 + 3x – 44 = 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 = 2 b = 3 c = – 44


= 2. – 44 = – 88


And either of their sum or difference = b


= 3


Thus the two terms are 11 and – 8


Sum = 11 – 8 = 3


Product = 11. – 8 = – 88


2x2 + 3x – 44 = 0


2x2 + 11x – 8x – 44 = 0


x(2x + 11) – 4(2x + 11) = 0


(2x + 11)(x – 4) = 0


x = 4 or – 11/2


x = 4 (x is a positive integers)


When x = 4


x + 1 = 4 + 1 = 5


x + 2 = 4 + 2 = 6


Hence the required integers are 4, 5, 6


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