A rectangular parking lot has a length that is 5 yards greater than the width. The area of the parking lot is 50 square yards. Find the length and the width.

Question

I need help on how to solve this equation.

Answer 1

Hi!

 

To solve this problem, you need to start by asking what you do and don't know (always where to start with word problems)

 

We know that Area (A) = Length (L) * Width (W)

 

You know Area (A) = 50 sq. yds.

 

You don't know anything about the width, so let's just say Width (W) = x

 

And you know that the length is 5 yds longer than the width, so L = W + 5yd and, since W = x, we can say L = x + 5yd

 

Now, we recall that A = L * W, and, since A = 50, we can say 50 sq yd = L * W.

 

Well, now just plug in. What does W equal and L equal? 

 

From above we plug in and can say that 50 sq yd = x * (x + 5 yd).

 

Multiply it out and you get 50 = x^2 + 5x (removing units for a moment so they don't clutter the lines).

 

Now there are a number of ways to solve this. You can tell (because of the "x^2") that it is quadratic. So let's get it equal to zero:

 

0 = x^2 + 5x - 50

 

Now you can factor it, use your quadratic formula, or (if you hated yourself) do it as a completing the square problem. I'm going to factor.

 

I can see from going through a list in my head that the factors 10 and -5 multiply to -50 and add to 5, so I know that

 

0 = x^2 + 5x - 50 = (x + 10)(x - 5)

 

And so you know that x = -10 or 5. 

 

Well, you know the parking lot cannot be -10 yd long. So we must say that W = x = 5, and, since L = W + 5, then L = 5+5 = 10. So your width is 5 yd and length is 10 yd.

 

Let me know if anything was unclear!