SOLUTION: Three consecutive even integers are such that twice the square of the third integer is 104 more than the product of the first two intergers. What are the three integers?
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Question 251112: Three consecutive even integers are such that twice the square of the third integer is 104 more than the product of the first two intergers. What are the three integers?
I just sent a word problem just like the one above, this is the revised problem. It should be "the first two" instead of "the first three". Sorry about the mix up and thanks in advance. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! LET X, X+2 & X+4 BE THE THREE INTEGERS.
2(X+4)^2=104+X(X+2)
2(X^2+8X+16=104+X^2+2X
2X^2+16X+32=104+X^2+2X
2X^2-X^2+16X-2X-104+32=0
X^2+14X-72=0
(X+18)(X-4)=0
X+18=0
X=-18 ANS.
X-4=0
X=4 ANS.
PROOF:
2(4+4)^2=104+4(4+2)
2*8^2=104+4*6
2*64=104=24
128=128
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2(-18+4)^2=104-18(-18+2)
2*-14^2=104-18*-16
2*196=104+288
392=392
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