Quadratic Word Problem

Question

The length of a rectangle is 1" more than the width. If 3" is added to the length and 1" to the width, the new area is twice the original area. Find the dimensions of the original rectangle

Answer 1

Let the dimensions of the original rectangle be x (length)and y (width).

The length is 1" more than the width, therefore

x = y + 1

The original area is

A1 = xy = (y+1)*y = y² + y

After adding 3" to the length and 1" to the width, the new dimensions become

length = x + 3 = (y+1) + 3 = y + 4

width = y + 1

The new area is

A2 = (y+4)(y+1) = y² + y + 4y + 4 = y² + 5y + 4

The new area is twice the original area.

Therefore,

A2 = 2*A1

y² + 5y + 4 = 2(y² + y)

y² + 5y + 4 = 2y² + 2y

Move all terms on the left to the right.

0 = y² - 3y - 4

This is a quadratic equation that can be factored as

(y + 1)(y - 4) = 0

The solutions are y = -1, or y = 4.

We cannot have negative length, therefore, ignore y = -1.

Because x = y+1, therefore x = 4+1 = 5

Answer:

The dimensions of the original rectangle are 5' (length) and 4" (width).

Check the answer:

5 = 4 +1 (the original length is 1" more than the width).

Original area = 5*4 = 20.

New area = (5+3)*(4+1) = 8*5 = 40, which is twice 20.