A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, How many feet of fencing will be required? 

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Find out how much fencing you need to enclose three sides of a rectangular field, leaving one side open at 20 feet. The field’s area is 680 square feet.

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, How many feet of fencing will be required?

The fencing required is 88 ft.

Here’s how

The problem can be solved by finding the dimensions of the rectangular field and then calculating the length of the fence required. Here’s how:

1. Find the length of the covered side:

   • We know the area of the field is 680 square feet and one side is 20 feet.
   • Area = Length x Width. So, Length = Area / Width = 680 sq. ft / 20 ft = 34 ft.

2. Calculate the total length of fencing required:

   • Since one side is already open (20 feet), we need to fence the remaining three sides.
   • The two covered sides have a total length of 2 * 34 ft = 68 ft.
   • Adding the length of the open side, the total fencing required is 68 ft + 20 ft = 88 ft.

Therefore, 88 feet of fencing will be required to enclose the rectangular field on three sides.

Quadratic Equations in Algebra

• A quadratic equation is a second-degree equation, meaning the highest power of the variable (usually denoted as x) is 2.
• It is typically written in the standard form: ax^2 + bx + c = 0, where a, b, and c are constants (a ≠ 0).
• These equations represent parabolas when graphed.

Solving Quadratic Equations:

There are several methods for solving quadratic equations, each with its own advantages depending on the specific equation:

1. Factoring:

• This method involves rewriting the equation as a product of two linear expressions using techniques like the sum-product pattern or grouping.
• Suitable when the equation can be easily factored (e.g., perfect squares, difference of squares).

2. Completing the Square:

• This method involves transforming the equation into a perfect square trinomial by adding and subtracting terms strategically.
• Useful when the equation is almost a perfect square but lacks a constant term.

3. Quadratic Formula:

• This is a general formula applicable to any quadratic equation: x = (-b ± √(b^2 – 4ac)) / 2a where a, b, and c are the coefficients in the standard form.
• While straightforward, it can be computationally intensive for complex equations.

4. Graphing:

• Plotting the parabola represented by the equation can visually locate the x-intercepts (solutions) where the graph intersects the x-axis.
• Provides an intuitive understanding of the equation’s behavior but may not yield precise solutions.

Additional Points:

• The number of solutions (roots) depends on the discriminant (b^2 – 4ac):
  • Two real solutions if the discriminant is positive (two x-intercepts).
  • One repeated real solution if the discriminant is zero (parabola touches the x-axis).
  • No real solutions if the discriminant is negative (parabola doesn’t touch or intersect the x-axis).
• Complex solutions (non-real numbers) can exist when the discriminant is negative, requiring complex number manipulations.

Examples:

• Factoring: Solve 2x^2 + 5x – 3 = 0. (Solution: x = -1, 3/2)
• Completing the Square: Solve x^2 + 4x + 1 = 0. (Solution: x = -2)
• Quadratic Formula: Solve 3x^2 – 7x + 2 = 0. (Solution: x = 2/3, 1)
• Graphing: Graph y = x^2 – 4x + 3 to find the solutions. (Solution: x = 1, 3)

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