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The length of a rectangle is \(3 \thinspace \rm{m}\) more than the
width. If the area of the rectangle is at least \(40 \thinspace
\rm{m}^2\), what are the possible dimensions of the rectangle?
Let \(x\) be the length. Then the width is \(x - 3\) and the area can be represented by \(A = x(x - 3)\). So, an inequality that represents this scenario is \(x(x - 3) \ge 40\).
Select a method to solve this inequality. Sign analysis is shown.
\(\begin{align}
x\left( {x - 3} \right) &\ge 40 \\
x\left( {x - 3} \right) - 40 &\ge 0 \\
x^2 - 3x - 40 &\ge 0 \\
\left( {x - 8} \right)\left( {x + 5} \right) &\ge 0
\end{align}\)
The inequality is true for \(x \le -5\) and \(x \ge 8\). However, because lengths must be positive, the interval \(x \ge 8\) represents the solution. The width would need to be greater than or equal to 5.
Select a method to solve this inequality. Sign analysis is shown.
\(\begin{align}
x\left( {x - 3} \right) &\ge 40 \\
x\left( {x - 3} \right) - 40 &\ge 0 \\
x^2 - 3x - 40 &\ge 0 \\
\left( {x - 8} \right)\left( {x + 5} \right) &\ge 0
\end{align}\)
The inequality is true for \(x \le -5\) and \(x \ge 8\). However, because lengths must be positive, the interval \(x \ge 8\) represents the solution. The width would need to be greater than or equal to 5.
| For more examples on solving quadratic inequalities in one variable, see pp. 476 – 483 of Pre-Calculus 11. |