The product of the first and twice the second is 40, what are the two integers?

2 Answers
Jul 26, 2015

Answer:

I found: #4 and 5# or #-5 and-4#

Explanation:

You can write (calling the first integer #n#):
#n*2(n+1)=40#
#2n^2+2n=40#
so:
#2n^2+2n-40=0#
Using the Quadratic Formula:
#n_(1,2)=(-2+-sqrt(4+320))/4=(-2+-sqrt(324))/4=(-2+-18)/4#
so:
#n_1=-5#
#n_2=4#

Jul 26, 2015

Answer:

If consecutive integers then #(4, 5)# or #(-5, -4)#, otherwise any pair of integers whose product is #20# will work.

Explanation:

If consecutive integers, then we are trying to solve:

#n * 2(n+1) = 40#

Divide both sides by #2# to get:

#n(n+1) = 20#

Subtract #20# from both sides and multiply out to get:

#0 = n^2+n-20 = (n-4)(n+5)#

So #n=4# or #n=-5#, meaning that the pairs of consecutive integers are:

#(4, 5)# or #(-5, -4)#

If the integers are not necessarily consecutive, then any integer pair of factors of #20# will work:

#(-20, -1)#, #(-10, -2)#, #(-5, -4)#, #(-4, -5)#,
#(-2, -10)#, #(-1, -20)#, #(1, 20)#, #(2, 10)#,
#(4, 5)#, #(5, 4)#, #(10, 2)#, #(20, 1)#