Question

The width of a rectangle is 12 feet shorter than twice its length. Find the width, and only the width, of the rectangle if the area is 224 square feet.

Answer 1

Ok, so the formula I keep ending up with is \[w =\frac{ 1 }{ 2 }L +4\]

Answer 2

but the equation \[134=L \left( \frac{ 1 }{ 2 }l +4 \right)\]

Answer 3

\[134=\frac{ 1 }{ 2 }L ^{2}\]

Answer 4

1/2 L^2 +4L

Answer 5

134=(4+(w/2))w

Answer 6

\[268=L ^{2}\]

Answer 7

you did't distribute!

Answer 8

\(134=L \left( \frac{ 1 }{ 2 }L +4 \right)= (1/2)L^2 +L\times 4= L^2/2+4L\)

Answer 9

you know there's a simpler way ...

Answer 10

let length =l and width = b
\[b=\frac{1}{2}l+4\]
Area of rectangle = lb
\[134=l(\frac{1}{2}l+4)\]
\[134=\frac{1}{2}l^2+4l \rightarrow 268 =l^2+8 l \rightarrow l^2+8 l -268=0\]
Now factorize it and u will find the ans

Answer 11

w= 1/2 L + 4
gives
L = (2w-4)

Area = Lw = (2w-4)w = 134
you directly get w from this

Answer 12

yeah, I ended up getting \[0=L ^{2}+8L -268\]
but I couldn't get that to factor

Answer 13

***2 (w-4)
Area = Lw = 2(w-4)w = 134

Answer 14

w^2-4w-68 = 0
got this ? or you want to continue with your work of finding L first ??

Answer 15

sec

Answer 16

Well now I got \[w ^{2}-4W -67=0\] so yes, i'm confused again

Answer 17

67 ? or 68 ?

Answer 18

that has irrational roots, not integer

Answer 19

yeah, its 67, sorry

Answer 20

We did several problems like this. It usually came out to a negative and positive integer. Since the length can't be negative, we just chose the positive one. So, I don't think I should have no solution for this, but i'm not sure. Math professors just love to give you wacky problems that they didn't go over.

Answer 21

w^2-4w-67=0
has 2 solutions, one positive and one negative.....

Answer 22

but both irrational

Answer 23

so i'm wondering whether I should say no rational numbers or just state the formula w=1/2L + 4

Answer 24

we don't deal with irrational or imaginary in this class.

Answer 25

you lost me after \[w ^{2}-4w -67=0\]

Answer 26

Compare your quadratic equation with \(ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two roots of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 27

heard of that quadratic formula ?

Answer 28

I know from mentally doing this equation that the answer is somewhere between 11 and 13

Answer 29

great!