find the dimensions

Question

The length of a rectangle is 4 meters less than twice the width. If the area of the rectangle is 286 square​ meters, find the dimensions.

Answer 1

Hi Mary

This problem combines two equations in two unknowns (l and w)

The length (l) is = to twice the width less 4. So l = 2w-4

Also, we know that the length X the width = 286 lw-286

Substitute l into the second equation (2w - 4)*w = 286, So 2w^2- 4 w = 286

2w^2 -4w - 286 = 0

Easiest way to address this is to divide the equation by 2 -------- w^2 -2w - 143 = 0

You can factor this ...... (w+13)(w-11)=0 Which gives values for wof -13 and 11. In real life situations like this, negative values for w can be rejected, because they don't make sense.

Thus w = 11. You can calculate l from either equation.

Answer 2

From the 1st sentence:

L = 2W - 4

From the 2nd sentence:

L * W = 286

Substitute the 1st equation for L in the 2nd:

(2W - 4) * W = 286

Expand and put in standard format:

2W² - 4W - 286 = 0

W² - 2W - 143 = 0

Factors:

(W - 13)(W + 11) = 0

W = 13, -11

Can't have a negative Width, so 13 is the answer.

L = 2(13) - 4 = 22

Check: 13 * 22 = 286