Question

the width of a rectangle is 9 less than twice its length. If the area of the rectangle is 64cm^2 what is the length of the diagonal?

Answer 1

We want to compute D, and we know that the area is

$\begin{gathered} x(2x-9)=64 \\ 2x^2-9x-64=0 \end{gathered}$

Then,

$\begin{gathered} x=\frac{9\pm\sqrt[]{(-9)^2-4(2)(-64)}}{2(2)} \\ x=\frac{9\pm\sqrt[]{593}}{4} \end{gathered}$

Since the length cannot be negative, then the length and with is

$\begin{gathered} length=\frac{9+\sqrt[]{593}}{4}=8.338 \\ \text{width}=2(\frac{9+\sqrt[]{593}}{4})-9 \\ \text{width}=\frac{-9+\sqrt[]{593}}{2}=7.676 \end{gathered}$

Now, using the Pythagorean theorem

$undefined$