SOLUTION: The length of a rectangle is 6 units less than the width. The area of the rectangle is 16 units. What is the length, in units, of the rectangle?

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 6 units less than the width. The area of the rectangle is 16 units. What is the length, in units, of the rectangle?       Log On


   

Question 845231: The length of a rectangle is 6 units less than the width. The area of the rectangle is 16 units. What is the length, in units, of the rectangle?

Answer by pmesler(52) About Me  (Show Source):
You can put this solution on YOUR website!
To start off let's write down the formulas that will be useful. We know that the length L is 6 units less than the width. We can write this
L = W - 6.
Next, let's write the formula for the area of a rectangle.
A = L * W.

Since we know the area we can plug that into the formula. We also know what the length L is. We can plug both of these into the formula.
16 = (W-6) * W
Now we simply solve for W.

16 = W^2-6w
This is starting to look like a quadratic equation. To make it a true quadratic equation, let's bring the 16 to the other side so the equation will equal zero.
W^2-6W - 16 = 0.
Now we simply use the quadratic equation to solve for W.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B-16+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-6%29%5E2-4%2A1%2A-16=100.

Discriminant d=100 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+100+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+100+%29%29%2F2%5C1+=+8
x%5B2%5D+=+%28-%28-6%29-sqrt%28+100+%29%29%2F2%5C1+=+-2

Quadratic expression 1x%5E2%2B-6x%2B-16 can be factored:
1x%5E2%2B-6x%2B-16+=+1%28x-8%29%2A%28x--2%29
Again, the answer is: 8, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B-16+%29


The solutions are x = -2 and x = 8. Obviously we can discard -2 as an extraneous root since you can't have a negative length. Therefore the width is 8 units.