Question #dc133

1 Answer
Oct 4, 2017

It is not clear that "The length of a rectangle is equal to 2 times the width plus 3 feet" is What?
So here Both the cases are solved.

When l=2b+3

Let the length of rectangle =l
and Width of rectangle =b

The length of a rectangle is equal to 2 times the width plus 3 feet.
Hence
l=2b+3 "feet" ...[A]

Area of Rectangle A=lb=20 "square feet"
Put the value of l from equation [A]
(2b+3)(b)=20
=>2b^2+3b-20=0
=>b=(-3+-sqrt((3^2)-4(2)(-20)))/(2xx2)=(-3+-sqrt(169))/4
=>b=(-3+-13)/4=-4 "or" 5/2
Negative value is discarded hence b=5/2
Hence l=2b+3
if l=2(5/2)+3=8

When l=2(b+3)

If
The length of a rectangle is equal to 2 times the width plus 3 feet.
Hence
l=2(b+3) "feet" ...[A]

Area of Rectangle A=lb=20 "square feet"
Put the value of l from equation [A]
2(b+3)(b)=20
=>b^2+3b=10
=>b^2+3b-10=0
=>b=(-3+-sqrt((3^2)-4(1)(-10)))/(2)=(-3+-sqrt(49))/4
=>b=(-3+-7)/4=1 "or" -5/2
Negative value is discarded hence b=1
Hence l=2b+3
if l=2(1)+3=5