Algebra - Applications of Quadratic Equations

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Section 2.8 : Applications of Quadratic Equations

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1. The width of a rectangle is 1 m less than twice the length. If the area of the rectangle is 100 m² what are the dimensions of the rectangle?

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We’ll start by letting L be the length of the rectangle. From the problem statement we now know that the width of the rectangle is 1 m less than twice the length and so must be \(2L - 1\) .

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We also know that the area of any rectangle is length times width and we are given that the area of this particular rectangle is 100. Therefore, the equation for this problem is,

\[\begin{align*}A & = \left( {{\mbox{length}}} \right)\left( {{\mbox{width}}} \right)\\ 100 & = \left( L \right)\left( {2L - 1} \right)\\ 100 & = 2{L^2} - L\end{align*}\] Show Step 3

This is a quadratic equation and we know how to solve that so let’s do that. First, we need to get the quadratic equation in standard form.

\[2{L^2} - L - 100 = 0\]

We can now use the quadratic formula on this to get,

\[L = \frac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 2 \right)\left( { - 100} \right)} }}{{2\left( 2 \right)}} = \frac{{1 \pm \sqrt {801} }}{4}\] Show Step 4

Reducing the two values we got in the previous steps to decimals we arrive at the following two solutions to the quadratic equation from Step 2.

\[L = \frac{{1 - \sqrt {801} }}{4} = - 6.8255\hspace{0.25in}\hspace{0.25in}L = \frac{{1 + \sqrt {801} }}{4} = 7.3255\]

We are dealing with a rectangle and so having a negative length doesn’t make much sense. Therefore the first solution to the quadratic equation can’t be the length of the rectangle.

This means that the length of the rectangle must be 7.3255 m and the width of the rectangle is then \(2\left( {7.3255} \right) - 1 = 13.651\,{\mbox{m}}\) .