The area of a square can be quadrupled by increasing the side length and width by 4 inches. What is the side length?

The area of a square is defined as $A=s^2$, where s is the side of the square.

Let x be the original side length of the square, then the area of the square with side x is $A=x^2$.

Now if the side lengths of the larger square are 4 inches longer than the side lengths of the original square, so, the side length of the new square will be $x+4$. And so, the area of the larger square is$A={(x+4)}^2$.

It is given that the larger square has an area that is 4 times the area of the original square so, the equation thus obtained will be ${(x+4)}^2=4x^2$.

Rewrite the equation ${(x+4)}^2=4x^2$in standard form.

$$\begin{gathered} {(x+4)}^2=4x^2 \\ (x+4)(x+4)=4x^2 \\ {x}^2+8x+16=4x^2 \\ -4x^2+x^2+8x+16=4x^2-4x^2\left[{\text{\hspace{0.17em}\hspace{0.17em}Subtract\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4x}}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}\hspace{0.17em}both\hspace{0.17em}\hspace{0.17em}sides\hspace{0.17em}\hspace{0.17em}}\right] \\ -3x^2+8x+16=0 \end{gathered}$$

Now solve for x by using the quadratic formula. Substitute $-3$for a, 8 for b, and 16 for c into the expression $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} x=\frac{-8\pm \sqrt{8^2-4\left(-3\right)\left(16\right)}}{2\left(-3\right)} \\ =\frac{-8\pm \sqrt{64+192}}{-6} \\ =\frac{-8\pm \sqrt{256}}{-6} \\ =\frac{-8\pm 16}{-6} \\ x=\frac{-8-16}{-6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{-8+16}{-6} \\ x=4\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{-8}{3} \end{gathered}$$

Since x represents the side length of the original square which must be a positive number, only $x=4$is a solution. Therefore, the side of the original square is then 4 inches.