Question

Three consecutive even integers are such that the product of the second integer and third integer is fifteen times the first integer. determine the three integers.

Answer 1

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Explanation:

Let's call the three integers

The product of the second and third integers is thus

We know that this quantity equals fifteen times the first integer:

The solutions of this equation are

So, the determinant has to be a perfect square.

This is not the case, because

which is not a perfect square.

So, it is not possible to find three consecutive integers with that property.

Answer 2

Let's call the three integers
`x,\ x+1,\ x+2`

The product of the second and third integers is thus

`(x+1)(x+2) = x^2+x+2x+2 = x^2+3x+2`

We know that this quantity equals fifteen times the first integer:

`x^2+3x+2 = 15x \iff x^2-12x+2=0`

The solutions of this equation are

`x_(1,2) = (-b\pm√(\Delta))/(2a)`

So, the determinant
`\Delta` has to be a perfect square.

This is not the case, because

`\Delta = b^2-4ac = 144-8 = 136`

which is not a perfect square.

So, it is not possible to find three consecutive integers with that property.