Question 159823
Let the size of the square piece of cardboard be x by x inches.
If you cut 2-inch squares from each of the four corners, the sides now measure (x-4) inches, and this will be the measure of the sides of the base of the newly-formed box.
The volume of a rectangular (s by s square base) prism with a height of h is given by:
{{{V = (s)(s)(h)}}} and, in this problem, s = (x-4) and h = 2, so...
{{{V = (x-4)(x-4)(2)}}} and this is to be 84.5 cu.in.
So we have enough information to set up the equation in x.
{{{(x-4)(x-4)(2) = 84.5}}} Performing the indicated multiplication, we get:
{{{2x^2-16x+32 = 84.5}}} Subtracting 84.5 from both sides, we have:
{{{2x^2-16x-52.5 = 0}}} Use the quadratic formula to solve this trinomial: {{{x = (-b+-sqrt(b^2-4ac))/2a}}} where: a = 2, b = -16, and c = -52.5.
Making the appropriate substitutions, we get:
{{{x = (-(-16)+-sqrt((-16)^2-4(2)(-52.5)))/2(2)}}} Simplifying this, we get:
{{{x = (16+-sqrt(256-(-420)))/4}}}
{{{x = (16+-sqrt(676))/4}}}
{{{x = 4+6.5}}} or {{{x = 4-6.5}}}
{{{x = 10.5}}} or {{{x = -2.5}}} Discard the negative solution as the sides of the box are positive, so we end up with:
{{{highlight(x = 10.5)}}}inches. This is the size of the original square piece of cardboard.
Let's check the solution:
{{{(x-4)(x-4)(2) = (10.5-4)(10.5-4)(2)}}}={{{(6.5)(6.5)(2) = (42.25)(2)}}} = {{{84.5}}} OK!