Algebra Word Problem 1

I've attached a visualization of the problem and  an abridged solution in the bottom file. 

Step 1. First, set the width of your rectangle equal to $x$ . Then, since we know that the length of the rectangle is 6 inches more than the width, we know that the length is equal to $x + 6$ . 

Step 2. We know that the area of a rectangle can be calculated by the following formula:
$Area = Length \ast Width$
By plugging in our length and width (in terms of x) we are able to create the following equation:
$x(x+6) = 16$
$x^2 + 6x - 16 = 0$

Step 3. In order to find x, we have to solve the quadratic equation, either through factoring or through the quadratic formula.

(Factoring Strategy)
If you choose to factor the quadratic equation, remember that we want to turn our equation into this form : $(x + p)(x + q) = 0$ , where p and q are any real numbers

Once we have our equation in this form, we know that the solutions to our equation will be $x = -p$ and $x = -q$. If you expand $(x + p)(x + q)$, you will have $x^2 + (p+q)x + pq$. If we compare this to our original $x^2 + 6x - 16 = 0$, we see that our product $pq$ will be equal to -16 and the sum of $p + q$ will be equal to 6. After some trial and error, you will find that p = -2 and q = 8 satisfy our requirements.

Then, we can factor $x^2 + 6x - 16 = 0$ into $(x-2)(x+8)= 0$, and we see that $x = 2$ and $x = -8$ are our solutions. However, since our width is equal to $x$ and width cannot be negative, we know that $x = -8$ is not a possible solution. Then, $x = 2$ is the only remaining solution and so, we know that our width has to be 2 inches. Thus, our length, which is equal to $x + 6$, will be 8 inches.  

(Quadratic Formula Strategy) 
For any quadratic equation $ax^2 + bx + c = 0$ where $a ≠ 0$ , the solutions will be $x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} $ as dictated by the quadratic formula. For our quadratic equation $x^2 + 6x - 16 = 0$,  we will have $a = 1, b = 6, c = -16$. We can plug this into the quadratic formula and get the following:

$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)} }{2(1)} = \frac{-6 \pm \sqrt{100} }{2} = \frac{-6\pm 10}{2}= {-3 \pm 5}$. Thus, we have $x = 2$ and $x = -8$. Remembering that our width is equal to x, we know that $x = -8$ is not a viable solution since width cannot be negative. As such, we know that $x = 2$ is the only possible solution and our width has to be 2 inches. Thus, our length, which is equal to $x + 6$, will be 8 inches.  

Final Answer. No matter if you decide to factor the quadratic equation or use the quadratic formula, you will arrive at the answer: width = 2 inches and length = 8 inches. Hope this helps and let me know if you have any questions.