ANSWER
[tex]width=7ft;\text{ }length=19ft[/tex]EXPLANATION
Let the length of the rectangle be L.
Let the width of the rectangle be W.
The length of the classroom is 2 feet shorter than three times its width. This implies that:
[tex]L=3W-2[/tex]The area of a rectangle is given by:
[tex]A=L*W[/tex]Hence, by substituting the given values into the equation, we have that:
[tex]\begin{gathered} A=(3W-2)*W \\ 133=3W^2-2W \\ \Rightarrow3W^2-2W-133=0 \end{gathered}[/tex]Solve the quadratic equation above by applying the quadratic formula:
[tex]W=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]where
a = 3, b = -2, c = -133
Therefore:
[tex]\begin{gathered} W=\frac{-(-2)\pm\sqrt{(-2)^2-(4*3*-133)}}{2(3)} \\ W=\frac{2\pm\sqrt{4+1596}}{6}=\frac{2\pm\sqrt{1600}}{6} \\ W=\frac{2\pm40}{6} \\ W=\frac{2+40}{6};\text{ }W=\frac{2-40}{6} \\ W=\frac{42}{6};\text{ }W=\frac{-38}{6} \\ W=7;\text{ }W=-\frac{19}{3} \end{gathered}[/tex]Since the width of a rectangle cannot be negative, the width of the rectangle is:
[tex]W=7\text{ }ft[/tex]To find the length of the rectangle, substitute the value of W into the equation for L:
[tex]\begin{gathered} L=3(7)-2=21-2 \\ L=19\text{ }ft \end{gathered}[/tex]Hence, the width of the rectangle is 7 ft and its length is 19 ft.
