The area of a rectangle is #20# #cm^2#. The length is #8# more than the width. How do you find the width in cm?

1 Answer
ILuvGaming101
Jun 11, 2017

The width of the rectangle is #2# cm.

Explanation:

Let's first recall the formula for the area of a rectangle:

#A=lw#

We can first substitute #20# into the area part of the equation since we already know its value:

#20=lw#

Now what? We have two variables but how do we find either of them? Looking back at the question, we see that "the length is 8 more than the width." Voila! Now we can write the length in terms of the width! So now we can switch out #l# for #w+8#:

#20=(w+8)w#

Now, using the Distributive Property, we can expand out the right side of the equation:

#20=w^2+8w#

We can subtract #20# from both sides to obtain a classic quadratic equation:

#w^2+8w-20=0#

Now, we can use factoring to simplify this problem. So our task now is to find two factors of #-20# such that they add up to #8#. We will list out all the factors of #-20# and then mark out the pair that adds up to #8# (I will use red for the pairs that don't work and blue for the pair that does work):

#color(red)(-1 and 20)#

#color(red)(1 and -20)#

#color(red)(-4 and 5)#

#color(red)(4 and -5)#

#color(blue)(-2 and 10)#

#color(blue)(2 and -10)#

We found our pair! 2 and -10 or -2 and 10! So now, we can write the quadratic equation in an #(w+a)(w+b)# format and make it equal to #0#:

#(w-2)(w+10)=0#

And since the product of #(w-2)# and #(w+10)# is 0, one of them must be equal to #0#. Let's start by assuming that #(w-2)# is equal to #0# and solve for #w#:

#(w-2)=0#
#w=2#

Now let's assume that #(w+10)# is equal to #0# and solve for #w#:

#(w+10)=0#
#w=-10#

Now, we can see that #w# could be either #-10# or #2#. However, there is no such thing as negative length so the possibility that #w=-10# is disqualified and #w=2# is the answer.

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