Question

PLEASE HELP!
The area of a rectangle is 52ft^2 , and the length of the rectangle is 5ft less than twice the width. Find the dimensions of the rectangle.

Answer 1

Let W be the width of the rectangle. It's unknown for now.

"the length of the rectangle is 5ft less than twice the width" tells us the length is

L = 2W - 5

Answer 2

so we have this rectangle

Answer 3

What is the area of any rectangle in general?

Answer 4

A=1/2*b*h

Answer 5

That's for a triangle

Answer 6

For a rectangle is

A = L*W

Answer 7

hahah yeah my bad

Answer 8

In this specific problem, this means

A = L*W

52 = (2W-5)*W

52 = W*(2W-5)

52 = 2W^2 - 5W

0 = 2W^2 - 5W - 52

2W^2 - 5W - 52 = 0

Now use the quadratic formula to solve for W

Answer 9

what quadratic formula?

Answer 10

Have you learned about this formula before?

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 11

yes, but how do i know which numbers to plug them into the letters

Answer 12

2W^2 - 5W - 52

is the same as

2x^2 - 5x - 52

after you replace W with x

Answer 13

2x^2 - 5x - 52 is in the form ax^2 + bx + c

a = 2
b = -5
c = -52

Answer 14

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(-52)}}{2(2)}\]

I'll let you finish