Question

The product of two consecutive negative integers is 1122. How do you find the integers?

Answer 1

`-34, and, -33.`

Explanation:

As we need two consecutive integers, let us suppose that, these

are `x and x+1," where, "x and (x+1) < 0.`

By what is given, `x(x+1)=1122.`

`:. x^2+x=1122.`

Multiplying by `4," we get, "4x^2+4x=4488.`

Completing the square, we have,

`4x^2+4x+1=4489.`

`(2x+1)^2=67^2.`

`2x+1=+-67.`

`because x < 0, &, (x+1) < 0; :. x+(x+1)=2x+1 < 0.`

`:. 2x+1=-67 rArr 2x=-67-1=-68.`

`:. x=-34, and, x+1=-33.`

Answer 2

Set up the equation `n xx (n+1) = 1122` solve for n and then find n+ 1

`-33xx -34 =1122`

Explanation:

`n xx (n + 1) = 1122` This gives

`n^2 + 1n = 1122` subtracting both sides by 1122 gives

`n^2 + 1n - 1122 = 1122 -1122` or

`n^2 + 1n - 1122 = 0`

`1122 = 33 xx 34`

`(n - 33) xx ( n + 34) = 0`

`n = 33 or n = -34`

Negative factors are asked for, so reject `n=33`

The factors are `-33 and -34`

Answer 3

The two integers are `-ceil(sqrt(1122)) = -34` and `-floor(sqrt(1122)) = -33`

Explanation:

If `0 < a < b` and `ab = n` then:

`a^2 < ab=n < b^2`

where all parts are positive. So taking the square root, we find:

`a < sqrt(n) < b`

So if `a` and `b` are consecutive positive integers then:

`a = floor(sqrt(n))" "` and `" "b = ceil(sqrt(n))`

For `n = 1122`, we find `sqrt(n) ~~ 33.496`

So:

`floor(sqrt(n)) = 33`

and

`ceil(sqrt(n)) = 34`

We find:

`33*34 = 1122`

as required.

Hence the negative solutions are `-34` and `-33` since:

`(-34)(-33) = 34*33 = 1122`

`color(white)()`
Hmmm - OK but how do you find `sqrt(1122)`?

All we need is a reasonable approximation for our purposes.

Start with:

`30^2 = 900 < 1122 < 1600 = 40^2`

So:

`30 < sqrt(1122) < 40`

Next, linearly interpolate:

`sqrt(1122) ~~ 30+(1122-900)/(1600-900)*10 = 30+222/700*10 ~~ 33`

Then we can use:

`sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+...)))`

with `a=33` and `b=1122-33^2 = 1122-1089 = 33`

So:

`sqrt(1122) = 33+33/(66+33/(66+33/(66+...)))`

which we can quickly see is about `33 1/2`, which is good enough for our purposes.

Or we could have simply noticed: Oh look, `1122-33^2 = 33` so `1122 = 33*34`, which is what we wanted to know.

Answer 4

`-34 and -33`

Explanation:

Note what happens when the factors of a number are arranged in order:

For example, `36`

`1,2,3,4,6,9,12,18,36`

The 'outer pair' `1 and 36`:
have the greatest sum `1+36=37`
and the greatest difference: `36-1=35`

The 'pair' in the middle would be `6 and 6`, but we only write one `6`
This gives the smallest sum, `6+6=12`
and the smallest difference: `6-6=0`

Also note that there is a single factor exactly in the middle because `36` is a perfect square and `sqrt36 =6`

A number such as `12` does not have a square root, but the middle factors are consecutive numbers, `3 and 4`

`12: 1,2,color(blue)(3,4),6,12`
`color(white)(xxxxxx)uarr`
`color(white)(xxxxxx)sqrt12`

`sqrt12` lies between `3and 4`

Therefore the same will happen with consecutive factors of `1122`, they will lie on either side of `sqrt1122`

`sqrt1122 =33.496`

The negative integers on either side are `-34 and -33`

Check: `-34xx-33 = 1122`