Chapter 4.7, Problem 39PPSE

Solution

a.

To write an equation using one variable to find the dimensions of the rectangle.

  $x^2-18x+36=0$

Given:

The perimeter of rectangle $=36\text{ in}$

The area of rectangle $=36{\text{ in}}^2$

Formula Used:

Perimeter of rectangle $=2\left(\text{length}+\text{width}\right)$

Area of rectangle $=\text{length}\times \text{width}$

Calculation:

Let the length of the rectangle $=x\text{ in}$

Now, the area of rectangle is:

  $\begin{array}{l}A=\text{length}\times \text{width}\\ \Rightarrow 36=\text{length}\times \text{width}\\ \Rightarrow 36=x\times \text{width}\\ \Rightarrow \text{width}=\frac{36}{x}\cdot \cdot \cdot \left(1\right)\end{array}$

Next, the perimeter of rectangle is:

  $\begin{array}{l}P=2\left(\text{length+width}\right)\\ \Rightarrow 36=2\left(\text{length+width}\right)\\ \Rightarrow \frac{36}{2}=x+\text{width}\\ \Rightarrow \text{width}=18-x\cdot \cdot \cdot \left(2\right)\end{array}$

Next, put the value of width from equation $(1)$ to equation $(2)$ to find the expression for $x$ :

  $\begin{array}{l}\frac{36}{x}=18-x\\ \Rightarrow x\left(18-x\right)=36\\ \Rightarrow 18x-x^2=36\\ \Rightarrow x^2-18x+36=0\end{array}$

b.

To check the discriminant and find the dimensions of rectangle.

  $15.72\text{ in and }2.28\text{ in}$

Given:

The equation is:

  $x^2-18x+36=0$

Formula Used:

Quadratic formula is given as:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Comparing the above equation with quadratic equation of the form $a{x}^2+bx+c,$ the parameters are:

  $\begin{array}{l}a=1\\ b=-18\\ c=36\end{array}$

The discriminant for the quadratic equation is:

  $\begin{array}{l}D=b^2-4ac\\ \Rightarrow D={\left(-18\right)}^2-4\left(1\right)\left(36\right)\\ \Rightarrow D=324-144\\ \Rightarrow D=180\end{array}$

Here, the discriminant is positive, hence there will be two roots for the quadratic equation. If the discriminant is negative, it means that there are no real roots for the quadratic equation.

Next, solving the quadratic equation as:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-\left(-18\right)\pm \sqrt{{\left(-18\right)}^2-4\left(1\right)\left(36\right)}}{2\left(1\right)}\\ \Rightarrow x=\frac{18\pm \sqrt{324-144}}{2}\\ \Rightarrow x=\frac{18\pm \sqrt{180}}{2}\\ \Rightarrow x=\frac{18\pm 6\sqrt{5}}{2}\\ \Rightarrow x=\frac{2\left(9\pm 3\sqrt{5}\right)}{2}\\ \Rightarrow x=9+3\sqrt{5}\text{ and }x=9-3\sqrt{5}\end{array}$

Hence, the values of $x$ are:

  $\begin{array}{l}x=9+3\sqrt{5}\\ \Rightarrow x=9+3\times 2.24\\ \Rightarrow x=9+6.72\\ \Rightarrow x=15.72\end{array}$

And

  $\begin{array}{l}x=9-3\sqrt{50}\\ \Rightarrow x=9-3\times 2.24\\ \Rightarrow x=9-6.72\\ \Rightarrow x=2.28\end{array}$

So, if the width of the rectangle is $15.72$ , then the length of the rectangle is $2.28$ . Similarly, if the width of the rectangle is $2.28$ , then the length of the rectangle is $15.72$ .

Hence, the dimensions of rectangle are $15.72\text{ in}$ and $2.28\text{ in}$ .