Each side of a square is increased 4 inches. When this happens, the area is multiplied by 4. How many inches in the side of the original square?

Let's denote the side length of the original square as \( x \). The area of the original square is \( x^2 \). When each side is increased by 4 inches, the new side length becomes \( x + 4 \), and the area of the new square is \( (x + 4)^2 \). According to the problem, this new area is four times the original area: \[ (x + 4)^2 = 4x^2 \] Expanding the left side: \[ x^2 + 8x + 16 = 4x^2 \] Now, rearranging the equation leads to: \[ 0 = 4x^2 - x^2 - 8x - 16 \] \[ 0 = 3x^2 - 8x - 16 \] Now, applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 3 \), \( b = -8 \), and \( c = -16 \). \[ b^2 - 4ac = (-8)^2 - 4 \cdot 3 \cdot (-16) = 64 + 192 = 256 \] Now, substituting back into the quadratic formula: \[ x = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6} \] This gives us two potential solutions: 1. \( x = \frac{24}{6} = 4 \) 2. \( x = \frac{-8}{6} = -\frac{4}{3} \) (not a viable solution since length can't be negative) Thus, the side of the original square is \( 4 \) inches. Now let's share some additional knowledge: The concept of squares and their areas has been around for centuries and is foundational in geometry. Ancient civilizations, including the Babylonians and Greeks, made extensive use of square shapes in their architecture and land division. The simplicity of squaring a number, like how we find areas, demonstrates the practical importance of this mathematical operation! If you're trying to solve similar problems in the future, remember to carefully translate the conditions given in the problem into mathematical equations. One common mistake is to misinterpret how the dimensions change—like confusing addition with multiplication in area-related questions. Always lay out your knowns and unknowns clearly before diving in!