Question

the length of a rectangle is 3 yd more than twice the width, and the area of the rectangle is 77 yd². find the dimensions of the rect length : yd width : yd

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yd. Then the length $l = 2w + 3$ yd.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. We know that $A = 77$ yd², so $(2w + 3)\times w=77$.
Expanding gives $2w^{2}+3w - 77 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b = 3$, $c=-77$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor.
Factor $2w^{2}+3w - 77=(2w + 11)(w - 7)=0$.
Setting each factor equal to zero:
$2w+11 = 0$ gives $w=-\frac{11}{2}$, but width cannot be negative.
$w - 7=0$ gives $w = 7$ yd.

Step4: Find the length

Substitute $w = 7$ into the length formula $l=2w + 3$.
$l=2\times7+3=14 + 3=17$ yd.

Answer:

Length: 17 yd
Width: 7 yd