Question

the width of a rectangle is 8 inches less than its length, and the area is 9 square inches. what are the length and width of the rectangle?

Answer 1

The given situation can be written in an algebraic way:

Say x the width of the rectangle and y its height.

- The width of a rectangle is 8 inches less than its length:

x = y - 8

- The area of the rectangle is 9 square inches:

xy = 9

In order to find the values of y and x, you first replace the expression

x = y - 8 into the expression xy = 9, just as follow:

`\begin{gathered} xy=9 \\ (y-8)y=9 \end{gathered}`

you apply distribution property, and order the equation in such a way that you obtain the general form of a quadratic equation:

`\begin{gathered} (y-8)y=9 \\ y^2-8y=9 \\ y^2-8y-9=0 \end{gathered}`

Next, you use the quadratic formula to solve the previous equation for y:

`y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

here you have a = 1, b = -8 and c = 9. By replacing these values you obtain:

`\begin{gathered} y=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(-9)}}{2(1)}=\frac{8\pm\sqrt[]{64+36}}{2} \\ y=\frac{8\pm\sqrt[]{100}}{2}=(8\pm10)/(2)=(8)/(2)\pm(10)/(2)=4\pm5 \end{gathered}`

Hence, you have two solutions for y:

y1 = 4 + 5 = 9

y2 = 4 - 5 = -1

You select only the positive solution, because negative lengths do not exist in real life. Hence, you have y = 9.

Finally, you replace the value of y into the expression x = y - 8 to obtain x:

`\begin{gathered} x=y-8 \\ x=9-8 \\ x=1 \end{gathered}`

Hence, the width and length of the given recgtangle are:

width = 1 in

length = 9 in