Question

The area of a rectangular garden is 54 square meters. The width is 7 meters longer than one-third of the length. Find the length and the width of the garden. Use the formula, are = length * width.

Answer 1

The width is 7 meters longer than one-third the length, so we can say

\[\Large W = \frac{1}{3}L + 7\]

Answer 2

The area of the rectangle is length * width, so

\[\Large A = L*W\]
\[\Large 54 = L*\left(\frac{1}{3}L + 7\right)\]

Do you see how to solve for L from here?

Answer 3

Yeah. Would it be \[54=\frac{ 1 }{ 3 }x^2+7x\]
From there I need help...

Answer 4

what I would do from there is multiply everything by 3 to get rid of the fraction

Answer 5

then get everything to one side

Answer 6

So like multiply 1/3x^2(3) and multiply 7x(3)?

Answer 7

yeah and 54 as well

Answer 8

Okie, one sec.

Answer 9

Got it. Would it be:
\[162=1x^2+21x\]

Answer 10

Or instead I should just put \[x^2\]

Answer 11

either works

Answer 12

now move that 162 over

Answer 13

Ok. I have a question... Does it matter with what number I multiply the numbers with?

Answer 14

you mean like how you multiplied everything by 3 just now?

Answer 15

Yeah.

Answer 16

The denominator of the fraction was 3, so multiplying everything by that clears out the denominator (ie eliminates the fraction)

Answer 17

if you picked another number that wasn't a multiple of 3, then the denominator or fraction wouldn't go away

Answer 18

Oh ok! Gotcha! Ok. So I got:
1x^2+21x-162=0
Give me a sec. I wanna do it myself and you tell me what I got wrong.

Answer 19

I got:
(x+54)(x-3)
PLEASE tell me it's right haha! I've had so much trouble with this today.

Answer 20

54 times -3 = -162 .... true

54 plus -3 = 21 ... false

so that's not the correct factorization

Answer 21

I'm confused and I think this is the stop. So how would I know which number goes where? Or more like why does it matter where I put down the sign with which number?

Answer 22

*spot

Answer 23

The quadratic formula may be easier to work with. Did you want to use that?

Answer 24

Yeah, I wanna use that instead... This is really hard for me and I have no idea why...

Answer 25

Ok, let me figure it out using the quadratic formula.

Answer 26

1x^2+21x-162

means

a = 1
b = 21
c = -162

plug those values into

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

Answer 27

I'm not exactly sure if this is right, but so far I have: \[x=\frac{ -21\pm \sqrt{207} }{ 2 }\]
Is that right?

Answer 28

*-201

Answer 29

The 207 portion is incorrect

b^2 - 4ac = 21^2 - 4*1*(-162) = 1089

Answer 30

So I multiplied it incorrectly? How?

Answer 31

21^2 = 441
4*1*(-162) = -648

Answer 32

21^2 - 4*1*(-162) = 441 - (-648) = 1089

Answer 33

Ahh! That's what I was missing! Gotcha! Ok. Hang on:)

Answer 34

So it'd be like \[x=\frac{ -21\pm \sqrt{1089} }{ 2 }\]
?

Answer 35

correct

Answer 36

Sorry for making you wait.... I'm kinda slow on thinking right now...

Answer 37

You're doing great. Don't worry.

Answer 38

I hope so... So my answer would be x=534?

Answer 39

The answer can never be negative.. I know that...

Answer 40

no

Answer 41

Ughgh. Ok, I need some real help. What is it?

Answer 42

\[\Large x=\frac{ -21\pm \sqrt{1089} }{ 2 }\]
\[\Large x=\frac{ -21+ \sqrt{1089} }{ 2 } \ \text{ or } \ x=\frac{ -21- \sqrt{1089} }{ 2 }\]
\[\Large x=\frac{ -21+ 33 }{ 2 } \ \text{ or } \ x=\frac{ -21- 33 }{ 2 }\]
\[\Large x=?? \ \text{ or } \ x=??\]

Answer 43

Wow, ok. I divided 1089 by 2 so that's what threw me off... Hang on let me get the final answer..

Answer 44

I got 6, -27.

Answer 45

x took the place of the length L

so L = 6 or L = -27

but a negative length makes no sense which is why we ignore -27

Answer 46

So, what would be my width? I don't seem to be understanding that part..

Answer 47

go back to \[\Large W = \frac{1}{3}L + 7\]

Answer 48

plug in L = 6

Answer 49

or you can ask yourself "something times 6 gives the area of 54, what is that something?"

Answer 50

Oh yeah. Sorry. I'm like dead tired and didn't get enough sleep lol.. 9

Answer 51

yep L = 6 and W = 9

Answer 52

Anyways, THANK YOU SO MUCH!!!!!! I can't appreciate you enough. I'd be sitting behind the computer sending you billions of messages with different thank you's. Thanks again!

Answer 53

I'm glad it's making sense now

Answer 54

Yeah, me too. I have another one similar to this, but I'll figure it out myself and let you check it since it's coming to my brain lol. Anyways, just expect a "check message" from me or something...

Answer 55

ok sounds like a good plan to me