Question 248681: Product of two consecutive integers is 89 more than their sum. Find integers
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! We know they are integers.
Let a and a+1 be the integers
a(a+1)=89+a+a+1
a^2+a=89+2a+1
a^2+a=90+2a
a^2-a=90
a^2-a-90=0
factor 90 that have a difference of 1
(9,10) done
(a+9)(a-10)=0
a=-9 (-9, -8)
and a=10 (10,11)
integers can be positive or negative
so we have two sets of possibilities.
we need to check them with the equation
a(a+1)=89+a+(a+1)
(-9)*(-8)=89+(-9)+(-8)
72=89-17
72=72
yes
let's try the positive numbers
10*11=89+10+11
110=110
yes
So we have two sets of integers
(-9,-8) and (9,10)
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