Chapter 9, Problem 44SGR

To determine

Find the side of the original square.

Answer to Problem 44SGR

The length of the side of the original square is 4 cm.

Explanation of Solution

Given:

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

Concept Used:

Let x be the original side length of the square. Write the expression for its area using the formula $A=s^2$ where s is the side length of the square.

Area of the square with side x is $A=x^2$

The side lengths of the larger square are 4 inches longer than the side lengths of the original square, so it has side length is $(x+4)$ .

The area of the larger square is $A={(x+4)}^2$

The larger square has an area that is 4 times the area of the original square so set the area of the larger square, ${(x+4)}^2$ equal to 4 times the area of the original square $x^2$

The equation is ${(x+4)}^2=4x^2$

Calculation:

The equation is ${(x+4)}^2=4x^2$

Rewrite the equation in standard form:

  $\begin{array}{l}$ {(x+4)}^2=4x^2$(x+4)(x+4)=4x^2\\ x^2+8x+16=4x^2\\ -4x^2+x^2+8x+16=4x^2-4x^2[\text{Subtract}{\text{4x}}^{\text{2}}\text{from}\text{both}\text{sides}]\\ -3x^2+8x+16=0\end{array}$

Now solve for x by using quadratic formula.

  $\begin{array}{l}a{x}^2+bx+c=0,\text{then}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ -3x^2+8x+16=0\\ a=-3;b=8;c=16\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-8\pm \sqrt{8^2-4·-3·16}}{2·-3}=\frac{-8\pm \sqrt{64+192}}{-6}=\frac{-8\pm \sqrt{256}}{-6}=\frac{-8\pm 16}{-6}\\ x=\frac{-8-16}{-6}\text{and}\frac{-8+16}{-6}\\ x=4\text{and}\frac{-8}{3}\end{array}$

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

The side of the original square is then 4 inches.

Thus, the length of the side of the original square is 4 cm.