Question

the length of a rectangle is 3 m less than double the width, and the area of the rectangle is 14 m². find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w\) meters. Then the length \( l=(2w - 3)\) meters.

Step2: Use the area formula

The area formula of a rectangle is \(A = l\times w\). Given \(A = 14\space m^{2}\), so \((2w - 3)\times w=14\).
Expand it: \(2w^{2}-3w - 14 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 2\), \(b=-3\), \(c = - 14\)), use the quadratic formula \(w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-3)^{2}-4\times2\times(-14)=9 + 112=121\).
Then \(w=\frac{3\pm\sqrt{121}}{4}=\frac{3\pm11}{4}\).
We get two solutions: \(w_{1}=\frac{3 + 11}{4}=\frac{14}{4}=3.5\) and \(w_{2}=\frac{3-11}{4}=\frac{-8}{4}=-2\).
Since the width cannot be negative, \(w = 3.5\space m\).

Step4: Calculate the length

Substitute \(w = 3.5\) into \(l=(2w - 3)\). Then \(l=2\times3.5-3=7 - 3=4\space m\).

Answer:

The width of the rectangle is \(3.5\space m\) and the length is \(4\space m\).