|
Question 954247: Good evening! I have two questions today:
1. Find three consecutive integers such that the product of all three, decreased by the cube of the first, is 33
2. Find three consectutive integers such that the product of the second and the third, decreased by four times the second, is five more than five times the first
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! x*(x+1)*(x+2)-x^3= 33 Reorder terms
x*(1+x)*(2+x)-x^3= 33 Multiply as follows:
(1+x)*(2+x) = 2+1x+2x+x^2 So we now have:
x(2+1x+2x+x^2)- x^3= 33 Add on left:
x(2+3x+x^2)- x^3= 33 Multiply to eliminate parenthesis
2x+3x^2+x^3- x^3 = 33 Simplify, subtract on the left
2x+3x^2 = 33 Now, to get a quadratic equation, subtract 33 on both sides
2x+3x^2-33 = 0 Factor the equation:
(x-3)(3x+11) = 0 Solve each side separately:
x-3= 0; x= 3
3x+11= 0; 3x= -11; x= -11/3
Discard the negative. Let's try the other answer which is 3:
3*(3+1)*3+2)-3^3=
3*4*5-3^3
60-3^3= 60-27= 33 We got the right answer: 3
|
|
|
| |
