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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.



Answer :

apologiabiology
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

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