The square of the first added to twice the second is 5, what are the two integers?

2 Answers
Jun 27, 2015

Answer:

There are an infinite number of solutions, the simplest and only positive integer solutions being 1 and 2.

Explanation:

For any #k in ZZ#

let #m=2k+1#
and #n=2-2k-2k^2#

Then:

#m^2 + 2n#

#= (2k+1)^2+2(2-2k-2k^2)#

#=4k^2+4k+1+4-4k-4k^2 = 5#

Jun 27, 2015

Answer:

If they are supposed to be consecutive integers, then the solution with negatives is the first is #-3# and the second is #-2#.
The positive solution is: first is #1# and second is #2#.

Explanation:

Assuming that these are supposed to be consecutive integers and the lesser integer is the first, then we can use:

first = #n# and second = #n+1#

The square of the first is #n^2# and twicwe the second is #2(n+1)#, so we get the equation:

#n^2+2(n+1) = 5#

(Note that this is not a linear equation. It is quadratic.)

Solve:

#n^2+2(n+1) = 5#
#n^2+2n+2 = 5#
#n^2 +2n-3=0#
#(n+3)(n-1) = 0#

#n+3=0# leads to #n=-3# and #n+1# = -2

If we check the answer, we get #(-3)^2+ 2(-2) = 9+(-4)=5#

#n-1=0# leads to #n=1# and #n+1# = 2

If we check this answer, we get #(1)^2+2(2) = 1+4 =5#