Questions on Algebra: Quadratic Equation answered by real tutors!

Algebra ->  Algebra -> Questions on Algebra: Quadratic Equation answered by real tutors!     (Log On)


Question 155598: The width of a rectangle is 2 ft less than the length. the area is 8 ft^2. Find the length and the width.
PLEASE HELP!!!: The width of a rectangle is 2 ft less than the length. the area is 8 ft^2. Find the length and the width.
PLEASE HELP!!!
Answer by jojo14344(807) About Me  (Show Source):
You can put this solution on YOUR website!
Remember: For rectangle,
Area=Length*Width----------------------------> working eqn
Given:
L=unknown=x
W=x-2ft
A=8ft^2
Substitute, 8=x*(x-2)
8=x^2-2x ------> x^2-2x-8=0
BY QUADRATIC: a=1, b=-2, & c=-8
x=(-b+-sqrt(b^2-4ac))/(2a)
x=(-(-2)+-sqrt(-2^2-4*1*-8))/(2*1)
x=(+2+-sqrt(4+32))/2
x=(2+-sqrt(36))/2
x=(2+-6)/2---> 2 values (2+6)/2=8/2=4,(2-6)/2=-4/2=-2
USED x=4ft ----------------> LENGTH
4-2=2ft -----------------> WIDTH
In doubt? Go back working eqn:
8ft^2=(4ft)(2ft)
8ft^2=8ft^2
Thank you,
Jojo