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?

1 Answer
Ben
Mar 12, 2018

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)#

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