Question

the area of a rectangle is 28 ft², and the length of the rectangle is 1 ft more than twice the width. find the dimensions of the rectangle. length : ft width : ft

Explanation:

Step1: Let the width be $x$ ft.

The length is $2x + 1$ ft. The area formula for a rectangle is $A=\text{length}\times\text{width}$. So we have the equation $x(2x + 1)=28$.

Step2: Expand the equation.

$2x^{2}+x = 28$, which can be rewritten as $2x^{2}+x - 28=0$.

Step3: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 1$, $c=-28$), we use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(1)^{2}-4\times2\times(-28)=1 + 224 = 225$. Then $x=\frac{-1\pm\sqrt{225}}{2\times2}=\frac{-1\pm15}{4}$. We get two solutions for $x$: $x_1=\frac{-1 + 15}{4}=\frac{14}{4}=3.5$ and $x_2=\frac{-1-15}{4}=\frac{-16}{4}=-4$. Since the width cannot be negative, we take $x = 3.5$ ft.

Step4: Find the length.

Substitute $x = 3.5$ into the length formula $l=2x + 1$. So $l=2\times3.5+1=7 + 1=8$ ft.

Answer:

Length : 8 ft
Width : 3.5 ft