Question

the length of a rectangle is 3 ft less than double the width, and the area of the rectangle is 65 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 of a rectangle is $A=l\times w$. Given $A = 65$ square - feet, we substitute $l$ and $A$ into the formula: $(2w - 3)\times w=65$.

Step3: Expand the equation

Expand $(2w - 3)w$ to get $2w^{2}-3w = 65$. Rearrange it to the standard quadratic - form $2w^{2}-3w - 65 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-3$, $c = - 65$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $2w^{2}-3w - 65 = 0$, we get $(2w + 13)(w - 5)=0$.

Step5: Find the value of $w$

Set each factor equal to zero:
If $2w+13 = 0$, then $2w=-13$, $w=-\frac{13}{2}$. But the width cannot be negative, so we discard this solution.
If $w - 5 = 0$, then $w = 5$ feet.

Step6: Find the value of $l$

Substitute $w = 5$ into the length formula $l = 2w-3$. So $l=2\times5 - 3=7$ feet.

Answer:

Length : 7 ft
Width : 5 ft