Question

find the dimensions of a rectangle whose width is 6 miles less than its length and whose area is 72square miles

Answer 1

Length = 12 miles

Width = 6 miles

Explanation:

Set up our two equations:

L * W = 72

W = L - 6

So, let's plug W in to find our values:

(Remember: W = L - 6)

L * (L - 6) = 72

Simplify:

L^2 - 6L = 72

L^2 - 6L - 72 = 0

Let's use the quadratic formula (a = 1, b = -6, c = -72)

Quadratic formula:

x = -b ±
`√(b^2-4ac)`

______________

2a

Plug in our values:

`(-\left(-6\right)\pm √(\left(-6\right)^2-4\cdot \:1\cdot \left(-72\right)))/(2\cdot \:1)`

Solve for the square root value:

`√((-6)^2-4*1*(-72)) = 18`

Set up the equation with simplified values:

L, W =
`(-\left(-6\right)\pm \:18)/(2\cdot \:1)`

`L_1=(-\left(-6\right)+18)/(2\cdot \:1),\:L_2=(-\left(-6\right)-18)/(2\cdot \:1)`

Solve for L1:
`(-\left(-6\right)+18)/(2\cdot \:1)` = 12

Solve for L2:
`(-\left(-6\right)-18)/(2\cdot \:1)` = 6

Since the width is 6 miles less than the length, the width is 6 miles, and the length is 12 miles.

Plug this in to make sure: Does 6 * 12 = 72? Yes. Is 12 6 more than 6? Yes. We have our answers

Length = 12 miles

Width = 6 miles