The length of a rectangle is 3 feet more than twice its width. The area is 65 square feet. Find the dimensions of the rectangle using the quadratic formula.

L=legnth

w=width

a=area

legnth is 3 more than twice its width

l = 3 + 2 times w

l=3+2w

a=65

a=l times w

subsitute

65=lw

l=3+2w

subsitute

65 = (3+2w) (w)

65=3w+2w^2

subtract 65 from both sides

0=2w^2+3w-65

if we can factor then we can do

if zy=0 then assume z and/or y=0 so

to factor in ax^2+bx+c form when a is greater than 1,

aw^2+bw+c

2=a

3=b

-65=c

to factor first find

a times c=z

b=t+y

t times y=z

so

a times c=2 times - 65=-130

b=3

factor - 130

-130=

-1,130

-2,65

-5,26

-10,13

now add them together and see which ones make 3 or - 3 (b)

-1+130=129

-2+65=63

-5+26=21

-10+13=3 match

seperate the middle number like that

3w=-10w+13w

so

2w^2-10w+13w-65=0

group

(2w^2-10w) + (13w-65) = 0

factor

(2w) (w-5) + (13) (w-5) = 0

reverse distribute ab+ac=a (b+c)

(2w+13) (w-5) = 0

set each to zero

2w+13=0

w-5=0

solve

2w+13=0

subtract 13

2w=-13

divide 2

w=-6 and 1/2

impossible since width cannot be negative discard

w-5=0

add 5 to both sides

w=5

subsitute

65=wl

65=5l

divide 5

13=l

width=5 feet

legnth=13 feet