Question

Question #dc133

Answer 1

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`