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Question
Three consecutive positive integers are such that the sum of the square of the first and product of the other two is 46. Find the integers.
Solution
Let the three consecutive positive integers be x, x +1and x + 2.
According to the given condition,
`x^2+(x+1)(x+2)=46`
⇒`x^2+x^2+3x+2=46`
⇒`2x^2+3x-44=0`
⇒`2x^2+11x-8x-44=0`
⇒`x(2x+11)-4(2x+11)=0`
⇒`(2x+11)(x-4)=0`
⇒`2x+11=0 or x-4=0`
⇒`x=-11/2 or x=4`
∴`x=4` ( x is a positive integer)
`x+1=4+1=5`
`x+2=4+2=6`
Hence, the required integers are 4, 5 and 6.
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