2. The area of a rectangle is 54cm. If the length is 3cm. longer than the width. What are the dimensions?

We use the steps for solving word problems. First define our unknowns. Here they are looking for two unknowns, the length and the width, but since we are given that the length is 3cm. longer than the width, if we know the width, we will be able to find the length. So

let   x = the width


x + 3 = the length

If we plug these into the formula for the area of a rectangle,

A = LW

we get

54 = (x + 3)x

Now that we have our equation, we run through the steps for solving equations. First, remove parentheses

54 = x2 + 3x

When we remove the parentheses, we see that we have a quadratic equation. So we run through the steps for solving quadratic equations. Transpose all terms to one side and leave a zero on the other.

0 = x2 + 3x - 54

Factor. This is one of the easier factoring problems. The   x squared   comes from an   x   and an   x   and we are then looking for two numbers that multiply to   54 Y  and have a difference of   3.   If you think about it, of course we are looking for two number that multiply to   54   and have a difference of   3.   The length and the width of the rectangle will multiply to   54   and have a difference of   3.   With thepractice we have been getting at factoring we soon realize that   6   and   9   are the numbers, and we will be done. But let's pretend that we didn't notice that. Factor

0 = (x - 6)(x + 9)

Set the factors equal to zero and solve the resulting equations.

This gives us two answers,   6   and   -9.   If the width is   6 cm.  then the length, being   3   cm more would be   9 cm,   and the area would be

(6cm)(9cm) = 54 sq cm.

which checks.

For the other solution,   -9 cm.,   it is difficult to imagine a width of   -9cm.  However, if you could have a width of   -9cm.,   then the length, which would be   3 cm.   more, would be   -6 cm.,   and the area would be

(-6cm)(-9cm) = 54 sq cm,

since a negative times a negative is a positive. However, the difficulty in imagining a negative width prompts most authors to say that distance is never negative, so the   -9   will not check because it is a distance. So when we are solving word problems, it is important to check the answers in the original equation because it is possible to get a solution which will check in the algebra but which will not make sense in the real world application.

The algebra doesn't know the situation, so it doesn't know that   -9   will not make sense. It will give you all of the answers whether a particular answer will make sense in the situation or not. Personally, I think that authors who ask their students not to consider what they call extraneous solutions take some of the fun out of it.

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