Question

1. the length of a rectangle is 1 foot more than twice the width. the area is 55 square feet. find the dimensions of the rectangle. let l = a = lw show set up, then solve. ( )( ) = 0 factor solve answer =

Explanation:

Step1: Define variables

Let $W$ be the width of the rectangle. Then the length $L = 2W + 1$. The area formula is $A=LW$, and $A = 55$.

Step2: Substitute variables into area formula

Substitute $L$ and $A$ into the formula: $55=(2W + 1)W$.

Step3: Expand the equation

Expand $(2W + 1)W$ to get $2W^{2}+W=55$, then rewrite it as a quadratic - equation $2W^{2}+W - 55=0$.

Step4: Factor the quadratic equation

Factor $2W^{2}+W - 55$: $2W^{2}+W - 55=2W^{2}+11W-10W - 55=W(2W + 11)-5(2W + 11)=(2W + 11)(W - 5)=0$.

Step5: Solve for $W$

Set each factor equal to zero:
If $2W+11 = 0$, then $2W=-11$, $W=-\frac{11}{2}$. But width cannot be negative, so we discard this solution.
If $W - 5=0$, then $W = 5$.

Step6: Find the length $L$

Substitute $W = 5$ into the length formula $L=2W + 1$. So $L=2\times5+1=11$.

Answer:

The width of the rectangle is 5 feet and the length is 11 feet.