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

then

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=x^2+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=x^2+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 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. So as soon as we realize that 6 and 9 are the numbers, 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.

 

x-6=0,x=6;x+9=0,x=9

 

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. However, it is difficult to imagine a width of -9cm. 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.

 

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