The product of two consecutive odd integers is 143. How do you find the integers?

Call x a number
The number (2x + 1) should be an odd number no matter what x is.
The consecutive odd number to (2x + 1) will be: (2x + 3).
There for:
(2x + 1)(2x + 3) = 143
#4x^2 + 6x + 2x + 3 = 143#
#4x^2 + 8x - 140 = 0#
#x^2 + 2x - 35 = 0#
Solve this quadratic equation for x. Find 2 numbers knowing sum (-2) and product (-35). They are x = 5 and x = -7 (rejected as negative)
The odd number is N = (2x + 1) = 10 + 1 = 11
The consecutive odd number is: N + 2 = (2x + 3) = 10 + 3 = 13
Their product is (11 x 13) = 143. OK

Let #n# be the even number between the two consecutive odd numbers #n-1# and #n+1#.

Then we have:

#143 = (n-1)(n+1) = n^2-1#

Add #1 # to both ends to get:

#144 = n^2#

So #n = +-sqrt(144) = +-12#

If #n = 12# then the two odd numbers are #11# and #13#.

If #n = -12# then the two odd numbers are #-13# and #-11#.

The consecutive odd numbers must lie on either side of the square root of 143

#sqrt 144 = 12# so #sqrt143 ~~ 11.9#

Look for odd numbers either side of #sqrt143#.

Try #11 and 13#

#11 xx 13 = 143" "larr# these are the factors we need.

Remember the factors could be negative as well..

#-11 xx -13 = 143#

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Background knowledge:
If you write the factors of a number in increasing order, note the following:

Factors are always in pairs.

• The difference between the factors is greatest for the outside factors and least for the innermost factors.

For example the factors of 48 are:

#1" "2" "3" "4" "6" "8" "12" "16" "24" "48#

48- 1 = 47 but 8-6 = 2

• The square root of a number is exactly in the middle of the factors

For example the factors of 36 are

#1" "2" "3" "4" "6" "9" "12" "18" "36#
#color(white)(.........................)uarr#
#color(white)(.......................)sqrt36

#6-6=0# this is the smallest possible difference.