9514 1404 393
Explanation:
This problem is demonstrating a solution to a word problem that involves the area of a rectangle and solution of a quadratic equations. We hope you learn from this example how to formulate and execute a strategy for solving word problems (or any other kind).
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<h3>Step 1</h3>
The problem is asking for dimensions, and tells you the relation of one dimension to another. It is generally convenient to assign a variable to a value the problem is asking for. Here, the unknown is the width of the garden. We choose to assign the variable "w" to remind us it stands for the width of the garden. The length units in the problem are meters, so we will assume that w stands for some number of meters.
The length of the garden is given as 12 m more than the width, so we can use the expression (w+12) to represent the length of the garden. The area of the garden is also given, and the shape is said to be a rectangle. This suggests we make use of the relation we know between length, width, and area for a rectangle.
A = LW
Using the expressions we have formulated, and the given value of the area, we can substitute into this formula to get ...
220 = (w+12)w
Why? Because this gives us an equation that we can solve to find the unknown dimensions of the garden that the problem asks for.
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<h3>Step 2</h3>
a) There are different possible strategies for solving the quadratic equation of Step 1. One of the easiest (after graphing with a graphing calculator) is "factoring." Most strategies for factoring start from the equation in "standard form." So, we put the equation in standard form based on our intent to solve it using factoring.
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b) The factored form of a quadratic equation is easy to solve. The solution can often be found mentally (as here). The numbers in the binomial factors are factors of the constant that have a sum equal to the coefficient of the linear term. Here, they are factors of -220 that have a sum of 12. Those factors are 22 and -10.
Often, these are found by searching through the possible factor pairs of -220 for a pair with the correct sum. Here, the most obvious factoring of 220 is 22·10, which factors differ by 12, so we have a useful clue immediately.
Why use factored form? It leads directly to a simple solution.
Why is this the factored form of the above equation? Because "simplifying" it gives the standard form equation.
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c) The zero product rule tells us a product will be 0 when one or more factors is zero. Here, we have the product of two binomial factors, and the equation tells us it is equal to zero.
Why are these the solutions? Each binomial will be zero when the value of the variable is the opposite of the value of the constant. We can mentally determine the opposite of the constant, so all we need to do is write down the possible solutions: the opposite of 22, and the opposite of -10.
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<h3>Step 3</h3>
a) The solutions we found are values of w, which we defined as the width of the garden. We know a width cannot be negative, so a value of -22 meters is not a solution to the problem.
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b) The other possible solution we found from the equation is w = 10. This corresponds to a width of 10 meters for the garden. There is nothing about this that would suggest it is anything other than a useful answer to the question. So, we conclude <em>the width of the garden is 10 meters</em>.
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c) The length of the garden is said to be 12 meters more than the width. That sum is 10 m + 12 m = 22 m. <em>The length of the garden</em> that satisfies the relations given in the problem statement <em>is 22 meters</em>.
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<em>Additional comment</em>
Various videos and websites describe the factoring of quadratic equations. It is easiest when the leading coefficient is 1, as here. You may notice that a negative sign on the constant in standard form means the factors have different signs.
Knowing that we will need factors of 220 that have a difference of 12, we can "cut to the chase" and solve the problem practically immediately by recognizing that 220 = 22·10, a pair of factors that differ by 12. This immediately tells us the dimensions of the garden are 10 m by 22 m. No further "work" is necessary.
