Question

the length of a rectangle is 3 ft less than double the width, and the area of the rectangle is 14 ft². find the dimensions of the rectangle. length : ft width : ft

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ feet. Then the length $l = 2w - 3$ feet.

Step2: Set up area - formula equation

The area formula for a rectangle is $A=l\times w$. Given $A = 14$, we substitute $l$ and $A$ into the formula: $14=(2w - 3)\times w$.

Step3: Expand the equation

Expand $(2w - 3)w$ to get $2w^{2}-3w=14$. Rearrange it to the standard quadratic - form $2w^{2}-3w - 14 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-3$, $c = - 14$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $2w^{2}-3w - 14=(2w - 7)(w + 2)=0$.

Step5: Find the value of $w$

Set each factor equal to zero:

• If $2w-7 = 0$, then $2w=7$, and $w=\frac{7}{2}=3.5$.
• If $w + 2=0$, then $w=-2$. But since the width cannot be negative, we discard $w=-2$.

Step6: Find the value of $l$

Substitute $w = 3.5$ into the length formula $l = 2w-3$. Then $l=2\times3.5-3=7 - 3=4$.

Answer:

Length : 4 ft
Width : 3.5 ft