From the square piece of cardboard, the length of the corners that were cut out would be:
[tex]\frac{5-x}{2}[/tex]
On folding the cardboard to form a box, the dimensions of the box now become:
[tex]\begin{gathered} Length(L)=x \\ Width(W)=x \\ Height(H)=\frac{5-x}{2} \end{gathered}[/tex]
If the volume of the box is to be 2 cubic inches, therefore,
[tex]\begin{gathered} Volume\text{ of box = }Length(L)\times Width(W)\times Height(H) \\ =x\times x\times\frac{5-x}{2} \\ =x^2(\frac{5-x}{2}) \\ \text{Equating the volumes,} \\ \frac{x^2(5-x)}{2}=2 \\ x^2(5-x)=4 \\ 5x^2-x^3=4 \\ R\text{earranging it,} \\ x^3-5x^2+4=0 \end{gathered}[/tex]
By trial and error method,
[tex]\begin{gathered} x=1\text{ is a solution} \\ \text{Hence, (}x-1)\text{ is one of the factors} \end{gathered}[/tex]
Using the polynomial long division method to find the other solution
The quotient obtained from the long division is
[tex]x^2-4x-4=0[/tex]
Using the quadratic formula to solve for x
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \text{where, }a=1,\text{ }b=-4,\text{ }c=-4 \\ x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-4)}}{2(1)} \\ x=\frac{4\pm\sqrt[]{16^{}+16}}{2(1)} \\ x=\frac{4\pm\sqrt[]{32}}{2} \\ x=\frac{4\pm4\sqrt[]{2}}{2} \\ x=2+2\sqrt[]{2}\text{ = }=2(1+\sqrt[]{2})\text{ =}4.828 \\ x=2-2\sqrt[]{2}\text{ = }=2(1-\sqrt[]{2})=-0.828 \end{gathered}[/tex]
Hence, all the rational solutions of the equation are:
[tex]\begin{gathered} x=1 \\ x=4.828 \\ x=-0.828 \\ \text{Note that a side length cannot be negative, so, }x\ne-0.828 \end{gathered}[/tex]
The possible side lengths of the box would be:
[tex]\begin{gathered} If\text{ }x=1 \\ Length(L)=1\text{ in} \\ Width(W)=1\text{ in} \\ Height(H)=\frac{5-1}{2}=\frac{4}{2}=2\text{ in} \end{gathered}[/tex][tex]\begin{gathered} If,x=4.828 \\ Length(L)=4.828\text{ in} \\ Width(W)=4.828\text{ in} \\ Height(H)=\frac{5-4.828}{2}=0.086\text{ in} \end{gathered}[/tex]
