Question

The area of a rectangle is 12 square inches. The length is 5 more than twice it’s width. How do you find the length and width?

Answer 1

Using the positive root in the quadratic equation, you find that `w=1.5`, which means `l=8`

Explanation:

We know two equations from the problem statement. First is that the area of the rectangle is 12:

`l*w=12`

where `l` is the length, and `w` is the width. The other equation is the relationship between `l` and `w`. It states that 'The length is 5 more than twice it's width'. This would translate to:

`l=2w+5`

Now, we substitute the length to width relationship into the area equation:

`(2w+5)*w=12`

If we expand the left-hand equation, and subtract 12 from both sides, we have the makings of a quadratic equation:

`2w^2+5w-12=0`

where:
`a=2`
`b=5`
`c=-12`

plug that into the quadratic equation:

`w=(-b+-sqrt(b^2-4ac))/(2a) rArr w=(-5+-sqrt(5^2-4(2*-12)))/(2*2)`

`w=(-5+-sqrt(25-(-96)))/4 rArr w=(-5+-sqrt(121))/4`

`w=(-5+-11)/4`

we know that the width must be a positive number, so we only worry about the positive root:

`w=(-5+11)/4 rArr w=6/4 rArr color(red)(w=1.5)`

now that we know the width (`w`), we can solve for the length (`l`):
`l=2w+5 rArr l=2(1.5)+5`

`l=3+5 rArr color(red)(l=8)`