SOLUTION: Find three consecutive even integers such that three times the sum of all three is thirty less than the product of the larger two integers.

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Question 711761:
Find three consecutive even integers such that three times the sum of all three is thirty less than the product of the larger two integers.
Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
LET X, X+2 & X+4 BE THE 3 INTEGERS.
3(X+X+2+X+4)+30=(X+2)(X+4)
3(3X+6)+30=X^2+2X+4X+8
9X+18+30-X^2-6X-8=0
-X^2+3X+40=0
X^2-3X-40=0
(X-8)(X+5)=0
X-8=0
X=8 ANS.
8+2=10 ANS.
8+4=12 ANS.
PROOF:
3(8+10+12)+30=10*12
3*30+30=120
90+30=120
120=120
X+5=0
X=-5 ANS.
-5+2=-3 ANS
-5+4=-1 ANS.
PROOF:
3(-5-3-1)+30=-3*-1
3(-9)+30=3
-27+30=3
3=3
PROOF:
3(-5-3-1)+30=-3*-1
3*-9+30=3
-27+30=3
3=3