Question

How do I solve this?

Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale).

A rectangle with dimensions x by x + 2 is inside a larger rectangle with dimensions 2x by x^2 + 5x + 6

Answer 1

can you write down the "formula" for the area of a rectangle that has sides x and x+2 ?

Answer 2

Would the formula for that be x over x + 2?

Answer 3

base * height

Answer 4

Oh so x * x + 2?

Answer 5

yes, but you need parens. it is x* (x+2)
if you write it x*x+2 that is x^2 +2 (which is not what you want.

Answer 6

next, what is the area for the big rectangle (same idea, different letters)

Answer 7

Okay
2x * (x^2 + 5x + 6)

Answer 8

ratio of small to big means write a fraction with the area of the small rectangle up top
and the area of the big rectangle in the bottom

Answer 9

we can simplify this ratio

Answer 10

so far we have
\[ \frac{x(x+2)}{2x(x^2 +5x+6)} \]

Answer 11

notice you can
1) "cancel" (or divide x/x) the x's
2) you can factor the quadratic.

can you factor x^2+5x+6 ?

Answer 12

Ummm... how would I do that?

Answer 13

Oh so it would end up in this format if I factored it right (x - )(x - )

Answer 14

something tells me you should have learned by now??

look at the last number (the +6)
list all ways to multiply 2 numbers to get 6:
1,6
2,3

(only two pairs... try that with 60 if you like to see lots of pairs)

the + on the +6 means the factors are the same sign.

Next, look at the +5 from +5x
the + means the biggest factor is + (and because of the "same sign" rule, both factors are +)

look for a pair that when they are both +, add up to +6

2,3 is the pair

that means the factors will be (x+2)(x+3)

Answer 15

I thought it would be (x - 2)(x - 3)

Answer 16

**look for a pair that when they are both +, add up to +5 <-- fixed typo

Answer 17

you can check by multiplying (use FOIL if that makes sense)
you will see that (x-2)(x-3)= x^2 -5x +6
you want x^2 + 5x + 6

Answer 18

Oooh okay that makes sense

Answer 19

once we have the factored form, you can simplify the ratio
what do we get ?

Answer 20

\[\frac{ x(x + 2) }{ 2x(x + 2)(x +3) }\]

Answer 21

See the Area of rect. is given by:

\[A = l \times b\]

So are of rect. one is

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

Area of bigger rect. 2:
\[2x(x^2 + 5x + 6) = = 2x^3 + 10x + 12x\]

No we know the ratio will be

\[\frac{ Area1 }{ Area2 }\]

\[\frac{ x(x +2) }{ 2x(x^2 +10x + 6) }= \frac{ x + 2 }{ 2x^2 + 10x + 6 }\]

No factoring out Area 2 =

\[2x^2 + 10x + 12 => 2x^2 + 6x + 4x + 12 => 2x(x + 3) + 4(x+ 3)\]

\[= (2x + 4 )(x+3)\]

\[=\frac{ x + 2 }{ (2x + 4 )(x+3)}\]

Answer 22

ok, now divide out any terms that are in the top and bottom (anything divided by itself is 1)

Answer 23

Can you cancel out the (x + 2)?

Answer 24

yes, (x+2)/(x+2) is 1 (no matter what x is *except* if x=-2 because that would be 0/0 )

so after "canceling" the (x+2) we should make a note: x≠ -2

Answer 25

and to be complete, because we also cancelled x/x we should make a note: x≠ 0

Answer 26

of course, a real rectangle does not have a side of length 0, so we will not be dividing by 0

Answer 27

After that would the fraction be

\[\frac{ x }{ 2x(x + 3) }\]

Answer 28

yes, but x/x is 1
so you can simplify some more.

Answer 29

I think I got it

Answer 30

\[\frac{ 1 }{ 2(x + 3) }\]

Answer 31

do you remember how to multiply fractions?
\[ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]
(multiply top times top and bottom times bottom
but we can "undo" that multiply:
\[ \frac{x}{2x(x+3)}= \frac{x}{x} \cdot \frac{1}{2(x+3)} \]
to make it more clear why it's ok to cancel the x/x (= 1)

Answer 32

and yes, you've got the answer

Answer 33

I'm sorry that took so long thank you so much for your help

Answer 34

yw

Answer 35

@Adjax too

Answer 36

so where I'm wrong guyz @phi and @kayla_renae

Answer 37

Idk @Adjax it just didn't come out the same and yours wasn't one of the answers Icould choose from

Answer 38

I wantmy answer to be reviewed as I wanted to know where I got wrong

Answer 39

Ask @phi what you did wrong

Answer 40

@Adjax did not do anything particularly wrong. but notice that
his answer can be simplified. Factor 2 from (2x+4):
\[ \frac{ x + 2 }{ (2x + 4 )(x+3)} \\\frac{ x + 2 }{ 2(x + 2 )(x+3)} \\
\frac{ \cancel{(x + 2)} }{ 2\cancel{(x + 2)}(x+3)} \\
\frac{1}{2(x+3)}
\]