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

4 Answers
Jul 22, 2017

# -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.#

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 #

Jul 22, 2017

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.

Jul 22, 2017

#-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#