Expressing Mathematical Sentences as Quadratic Equations: An Example with Rectangles

When dealing with geometric shapes, particularly rectangles, it’s often necessary to express certain relationships in the form of mathematical sentences. This article will take you through the process of expressing a given statement as a quadratic equation. We will show you how to represent these expressions in terms of a variable and solve for the unknown quantity.

Understanding the Problem

The given problem deals with a rectangle, where the length is six units less than twice the width. The area of the rectangle is 36 square units. We need to express this in a quadratic equation and solve for the width and subsequently the length of the rectangle.

Step-by-Step Solution

1. **Define the Variables:** - Let ( w ) represent the width of the rectangle. - Then, the length ( l ) can be expressed as ( l 2w - 6 ).

2. **Formulate the Area Equation:** - The area of a rectangle is given by the formula: ( A l times w ). - Substituting the expressions for ( l ) and ( A ), we get:

[begin{align*} 36 (2w - 6) times w 36 2w^2 - 6w end{align*}]

3. **Rearrange the Equation:** - To get the equation in standard quadratic form, we move all terms to one side:

[[0 2w^2 - 6w - 36]]

4. **Simplify the Equation:** - Divide all terms by 2 for simplicity:

[[0 w^2 - 3w - 18]]

Expressing the Quadratic Equation in Terms of ( x )

Since the problem involves a rectangle, we can replace ( w ) with ( x ). Therefore, the quadratic expression in terms of ( x ) is:

[x^2 - 3x - 18 0]

This is the final quadratic expression, and it is now in terms of ( x ).

Solving the Quadratic Equation

To solve for ( x ), we can factor the quadratic expression:

[[x^2 - 3x - 18 (x - 6)(x 3) 0]]

This gives us two possible solutions for ( x ):

[[x - 6 0 Rightarrow x 6]] [[x 3 0 Rightarrow x -3]]

Since ( x ) represents the width of a rectangle, it cannot be negative. Therefore, ( x 6 ).

Calculating the Length

Given ( x 6 ) (or ( w 6 )), we can find the length ( l ) using the relation ( l 2w - 6 ):

[[l 2 times 6 - 6 12 - 6 6]]

Thus, both the width and length of the rectangle are 6 units.

Conclusion

We have successfully expressed the given problem as a quadratic equation, solved it, and found that the rectangle is a square with both dimensions measuring 6 units. Understanding how to express and solve such equations is crucial in many areas of mathematics, including geometry and algebra.

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