Question

The two consecutive positive integers has a product of 272? What are the 4 integers?

Answer 1

`(-17,-16)` and `(16,17)`

Explanation:

Let a be the smaller of the two integers and let a+1 be the bigger of the two integers:
`(a)(a+1) = 272`, easiest way to solve this is to take the square root of 272 and round down:
`sqrt(272) = \pm16...`
16*17 = 272
Thus, the integers are -17,-16 and 16,17

Answer 2

16 17

Explanation:

If we multiply two consecutive numbers, `n and n+1`
we get `n^2+n`. That is we square a number and add one more on.

`16^2=256`

256+16=272

So our two numbers are 16 and 17

Answer 3

16 and 17

Explanation:

`color(blue)("A sort of cheat way")`

The two number are very close to each other so lets 'fudge' it

`sqrt(272) = 16.49...` so the first number is close to 16

Test `16xx17=272 color(red)(larr"First guess gets the prize!")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The systematic way")`

Let the first value be `n` then the next value is `n+1`

The product is `n(n+1)=272`

`n^2+n-272=0`

Compare to: `ax^2+bx+c=0color(white)("ddd") -> color(white)("ddd") x=(-b+-sqrt(b^2-4ac))/(2a)`
In this case `x->n; color(white)("d")a=1; color(white)("d")b=1 and c=-272`

`n=(-1+-sqrt(1-4(1)(-272)))/(2(1))`

`n=-1/2+-sqrt(1089)/2`

`n=-1/2+-33/2` The negative is not logical so discard it

`n=-1/2+33/2 = 16`

The first number is 16 the second is 17