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find the 3 consecutive positive integers such that eight times the sum of all three equals the product of the larger two. |
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Answer» The first integer is X The SECOND integer is X+1 The third integer is X+2
Three times the sum of all three....
3( X + X+1 + X+2) = 3( 3X + 3) = 9X + 9
the product of the larger two.... (X+1)(X+2) = X^2 + 3x + 2
The PROBLEM says those expressions are equal...
9X+9 = X^2 + 3X + 2 0 = X^2 + 3X + 2 - 9X - 9
0 = X^2 - 6X - 7
0 = ( X - 7)(X + 1)
X-7 = 0 ---> X = 7; So 7, 8, and 9
3( 7 + 8 +9) = 3(24) = 72 = 8 * 9 <--- yes it works!!
X+1=0 ---> x = -1; So -1, 0. and 1
3( -1 + 0 +1) = 3(0) = 0 = 0*1 <--- yes it works!!!
So there are 2 sets of consecutive integers that satisfy these CONDITIONS: {7,8,9} and {-1,0,1} |
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