If two consecutive positive integers have the property that one integer times twice the other equals 612 what is the sum of these two integers?

1 Answer
May 31, 2015

If the smaller integer is #x#
we have either #(x) xx 2(x+1) = 612# or #2(x) xx (x+1) = 612#

in either case
#color(white)("XXXXX")##2(x)(x+1) =612#

#color(white)("XXXXX")#2x^2+2x = 612#

#color(white)("XXXXX")#2x^2+2x-612 = 0#

Using the quadratic formula
#color(white)("XXXXX")##x = (-2+-sqrt(4+4(2)(612)))/(2(2)#

#color(white)("XXXXX")##=(-2+-sqrt(4900))/4#

#color(white)("XXXXX")##=(-2+-70)/4#
(ignoring the negative value, since we are told the numbers are positive)
#color(white)("XXXXX")##x=68/4 = 17#

and the two consecutive positive integers are 17 and 18