SOLUTION: find two consecutive even integers whose product is 80

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Question 445230: find two consecutive even integers whose product is 80
Found 2 solutions by mananth, unlockmath:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let the integers be x & x+2
x(x+2)=80
x^2+2x-80=0
x^2+10x-8x-80=0
x(x+10)-8(x+10)=0
(x+10)(x-8)=0
x= 8 OR -10
The numbers are
8 & 10
OR
-10,-8

Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
Let's have x be one even integer and (x+2) be the next. Now we can make and equation as:
x(x+2)=80
Expand out and subtract 80 from both sides to get:
x^2+2x-80=0
Factor this:
(x-8)(x+10)=0
Solve for x:
x=8
x=-10
Two answers: 8 &10
or -10 &-8
There we go.
Make sense?
RJ
www.math-unlock.com