Question

A den is 8 feet longer that it is wide. if the dens area is 308 square feet what are the dimensions of the room

Answer 1

this being algebra, you should choose letters of the alphabet to represent the width and then the length of the den.

Let the width be ... ?

Let the length be ... ? show how the length is related to the width.

Answer 2

how can i set up the equation?

Answer 3

Representation of your unknowns comes before setting up any equation.

Choose a letter, now, to represent the width of the den.

Answer 4

w for width
l for length

Answer 5

Let x = width of den.
Since the length is 8 feet longer than the width, the length is represented by x + 8.

Given the L and the W of any rectangle, how would you find the area?

Write out the area formula, please.

Answer 6

A=WL
A=X*X+8
Is that right?

Answer 7

Yes, but you do have to enclose that "x+8" inside parentheses.

Do that now, please. A =

Answer 8

A=x+(x+8)

Answer 9

But you multiplied in your first try, but now are adding. Area is L * W, not L + W.

Try writing the equation again, please. A = 308 sq ft =

Answer 10

A=308 sq ft= x*(x+8)

Answer 11

good. Please multiply out the right hand side of this equation.

Answer 12

x^2+8x

Answer 13

x^2+8x=308. Good. How would y ou solve this quadratic equation?

Answer 14

subtract 308 from both sides

Answer 15

yes

Answer 16

I'm going to use quadratic formula, hold on please

Answer 17

does x=14, -22?

Answer 18

-22 could not be correct, because length is never negative. That leaves you with 14.

Check your work.

x(x+8) must equal 308.

Substitute 14 for x. Is the resulting equatin true or false?

Answer 19

ok one second

Answer 20

yes i got 308

Answer 21

Very good. Great. That means your work is correct and you've solved this algebra problem. Keep in mind the importance of being able to represent unknown quantities with letters of the alphabet, as we did with W = x and L = x + 8.

Over and out, good night!

Answer 22

thank you so much

Answer 23

You're welcome, Daisy! Bye!