Question

Please check my work?
Rectangle A has an area of 4 − x^2. Rectangle B has an area of x^2 + 2x − 8. In simplest form, what is the ratio of the area of Rectangle A to the area of Rectangle B? Show your work.

Rectangle A
4 - x^2
(x + 2)(x - 2)

Rectangle B
x^2 + 2x - 8
(x + 4)(x - 2)

(x + 2)(x - 2)/(x + 4)(x - 2)
x + 2/x + 4

The ratio of the area of Rectangle A to the area of Rectangle B is x+2/x+4.

Is it correct?

Answer 1

Incorrect

Answer 2

nope your factoring of rectangle A is incorrect.

is the area is 4 - x^2 it factors to (2 - x)(2 + x)

Answer 3

4 - x^2 = (2+x)(2-x), NOT (x-2)(x+2)

Answer 4

Rectangle A
4 - x^2
(2 + x)(2 - x)

Rectangle B
x^2 + 2x - 8
(x + 4)(x - 2)

(2 + x)(2 - x)/(x + 4)(x - 2)
2 + x/x + 4

The ratio of the area of Rectangle A to the area of Rectangle B is 2+x/x+4.

is this right?

Answer 5

You have factored 4 - x^2 incorrectly. It should be (2 -x)(2 + x)

Answer 6

nope the scaling factor is negative

(2 - x) and (x - 2) are not the same

-(x - 2) can be cancelled with (x -2) but its -1

Answer 7

Sorry, you are right, I didnt see that you factored it correctly.

Answer 8

It cancels with a -1

Answer 9

Where do I get the -1 though?

Answer 10

so you are looking at

\[\frac{-(x -2)(x+2)}{(x-2)(x+4)}\]

now remove the common factor

the initial area equation for rectangle A can be written as

\[-(x^2 - 4) = -(x-2)(x+2)\]

Answer 11

I'm sorry... I still don't get it...

Answer 12

What I (and others) are saying: the expression (x-2)/(2-x) or (2-x)/(x-2) = -1

Answer 13

2-x and x-2 are NOT identical, so you just cant cancel them out so fast; but they will cancel as the expression equals - 1.

Answer 14

x+ 2 and 2+ x are identical, so they cancel out.

Answer 15

I get that part I think, I just don't know how to write this all out

Answer 16

start from the basics

write the ratio

you need the rewrite the numerator so that you can get a common factor.

\[\frac{4 - x^2}{x^2 + 2x - 8} = \frac{-(x^2 - 4)}{(x -2)(x + 4)}\]

Answer 17

which when factored is

\[\frac{-1(x -2)(x+2)}{(x-2)(x+4)}\]

only now can you cancel a common factor

which leaves

\[\frac{-(x +2)}{(x + 4)}\]

Answer 18

Campbell is correct.

Answer 19

Rectangle A
4 - x^2
-(x^2 - 4)
-(x-2)(x + 2)

Rectangle B
x^2 + 2x - 8
(x + 4)(x - 2)

-(x - 2)(x + 2)/(x + 4)(x - 2)
x + 2/x + 4

The ratio of the area of Rectangle A to the area of Rectangle B is x+2/x+4.

Is this one right?

Answer 20

-(x+2)/(x+4)

Answer 21

Kind of strange to have a negative ratio in the context of this problem.

Answer 22

But thats what it is algebraically.

Answer 23

Assuming that you typed the original question correctly.

Answer 24

Rectangle A
4 - x^2
-(x^2 - 4)
-(x-2)(x + 2)

Rectangle B
x^2 + 2x - 8
(x + 4)(x - 2)

-(x - 2)(x + 2)/(x + 4)(x - 2)
-(x + 2)/(x + 4)

The ratio of the area of Rectangle A to the area of Rectangle B is -(x+2)/(x+4).

How about this one??

Answer 25

I agree with that.

Answer 26

it's right??

Answer 27

I said, that I would agree with that.

Answer 28

Which means that you are right.

Answer 29

haha okay! Well thank you for everything!!

Answer 30

No problem.