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.

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) (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
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