Question

the length of a rectangle is 5 yd less than three times the width, and the area of the rectangle is 50 yd². find the dimensions of the rectangle.

Explanation:

Step1: Set up variables

Let the width of the rectangle be \( w \) yd. Then the length \( l=(3w - 5) \) yd.

Step2: Use the area formula

The area formula for a rectangle is \( A=l\times w \). Given \( A = 50\) yd², so \( (3w - 5)\times w=50 \).
Expand to get \( 3w^{2}-5w - 50=0 \).

Step3: Solve the quadratic equation

For a quadratic equation \( ax^{2}+bx + c = 0\) (\(a = 3\), \(b=-5\), \(c = - 50\)), use the quadratic formula \( w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-5)^{2}-4\times3\times(-50)=25 + 600=625\).
Then \( w=\frac{5\pm\sqrt{625}}{6}=\frac{5\pm25}{6}\).
We have two solutions for \( w\):
\( w_{1}=\frac{5 + 25}{6}=\frac{30}{6}=5\) and \( w_{2}=\frac{5-25}{6}=\frac{-20}{6}=-\frac{10}{3}\).
Since width cannot be negative, \( w = 5\) yd.

Step4: Find the length

Substitute \( w = 5\) into \( l=(3w - 5)\). Then \( l=3\times5-5=15 - 5=10\) yd.

Answer:

Length: \( 10\) yd, Width: \( 5\) yd