Question

the length of a rectangle is 5 m more than twice the width, and the area of the rectangle is 42 m². find the dimensions of the rectangle.
length:
width:

Explanation:

Step1: Set up variables

Let the width of the rectangle be \(x\) meters. Then the length is \((2x + 5)\) meters.

Step2: Use the area formula

The area of a rectangle is \(A=\text{length}\times\text{width}\). So, \(x(2x + 5)=42\).
Expand to get \(2x^{2}+5x-42 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 5\), \(c=-42\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(5)^{2}-4\times2\times(-42)=25 + 336=361\).
Then \(x=\frac{-5\pm\sqrt{361}}{2\times2}=\frac{-5\pm19}{4}\).
We have two solutions for \(x\):
\(x_{1}=\frac{-5 + 19}{4}=\frac{14}{4}=3.5\) and \(x_{2}=\frac{-5-19}{4}=\frac{-24}{4}=-6\).
Since the width cannot be negative, we take \(x = 3.5\).

Step4: Find the length

Substitute \(x = 3.5\) into the length formula \(l=2x + 5\).
\(l=2\times3.5+5=7 + 5=12\).

Answer:

Length: \(12\) m, Width: \(3.5\) m