A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.
You should know that the formula for the volume of a rectangular prism is
V = LWH.
So plug in:
$\displaystyle 504 = x(x + 1)(x + 2)$
(let's switch sides)
$\displaystyle x(x + 1)(x + 2) = 504$
Multiply out the left side, and afterwards, subtract 504 from both sides. You'll end up with a cubic equation equaling zero. Solve for x. Don't forget to reject any solution that gives you a negative number for any of the lengths.
01
EDIT: beaten to it by skeeter!
first of all, correct your expansion ...
$\displaystyle x(x+1)(x+2) = 504$
$\displaystyle x^3 + 3x^2 + 2x - 504 = 0$
you know 7 is a root ... use synthetic division
note the coefficients of the depressed polynomial ...Code:7] 1.......3........2.......-504 ............7.......70........504 ------------------------------- ....1......10.......72.........0
$\displaystyle x^2 + 10x + 72 = 0$
since $\displaystyle b^2-4ac < 0$ , the other two roots are imaginary.
$\displaystyle x = 7$ is the only real root
go back to the original equation ...
$\displaystyle x(x+1)(x+2) = 504$
$\displaystyle 7(7+1)(7+2) = 504$
$\displaystyle x(x+1)(x+2) = 504$
$\displaystyle x(x^2+3x+2)=504$
$\displaystyle x^3+3x^2+2x-504=0$
$\displaystyle (x-7)(x^2+10x+72)=0$
Notice that there are no real solutions for the quadratic factor as $\displaystyle \triangle <0$
So the only real solution is $\displaystyle x=7$
The question stated the sides are $\displaystyle x,\ x+1,\ x+2$ so one side is $\displaystyle 7$ and the others are $\displaystyle 7+1=8$ and $\displaystyle 7+2=9$