Question 1075927: The sum of the squares of two consecutive even numbers is 84 more than the square of the next larger even number. What are the numbers? Found 2 solutions by jorel1380, ankor@dixie-net.com:Answer by jorel1380(3719) (Show Source):
You can put this solution on YOUR website! Let n be the first even number. Then the next two even numbers are n+2 and n+4. So:
n²+(n+2)²-84=(n+4)²
n²+n²+4n+4-84=n²+8n+16
n²-4n-96=0
(n+8)(n-12)=0
n=-8 or 12
We get a positive value for n of 12; then n+2 and n+4 are 14 and 16, respectively. ☺☺☺☺
You can put this solution on YOUR website! Let the two numbers be n and (n+2), the next larger even number is (n+4)
Write an equation for exactly what it says
"the sum of the squares of two consecutive even numbers is 84 more than the square of the next larger even number."
n^2 + (n+2)^2 = (n+4)^2 + 84
FOIL
n^2 + n^2 + 4n + 4 = n^2 + 8n + 16 + 84
Combine like terms on the left
n^2 + n^2 - n^2 + 4n - 8n + 4 - 16 - 84 = 0
n^2 - 4n - 96 = 0
we know these are integers so this will factor
(n-12)(n+8) = 0
The positive solution is what we want here
n = 12
:
What are the numbers? 12 and 14
:
;
Check this on your calc
12^2 + 14^2 =
16^2 + 84 =
:
:
Note that the solution n = -8 will also satisfy the given requirement
-8^2 + -6^2 = -4^2 + 84
64 + 36 = 16 + 84
