Question

find the 3 consecutive positive integers such that eight times the sum of all three equals the product of the larger two.

Answer 1

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}

Answer 2

Answer:

sorry I couldn't understand