Question

Write a quadratic equation and solve using any algebraic method. The area of a rectangular carpet is 216 square feet. The length of the carpet is 3 feet more than three times the width of the carpet. What are the dimensions (length and width) of the carpet?

Answer 1

Let x be the width of the carpet, then the length of the carpet is 3 + 3x.The area of the carpet is the product of the length and the width, so we have:x(3 + 3x) = 216Expanding the brackets and simplifying, we get:3x^2 + 3x - 216 = 0This is a quadratic equation in standard form ax^2 + bx + c = 0, where a = 3, b = 3, and c = -216.To solve this equation, we can use the quadratic formula:x = (-b ± sqrt(b^2 - 4ac)) / 2aSubstituting the values of a, b, and c, we get:x = (-3 ± sqrt(3^2 - 4×3×(-216))) / 2×3

x = (-3 ± sqrt(9 + 2592)) / 6

x = (-3 ± sqrt(2601)) / 6x = (-3 ± 51) / 6x = 8 or x = -9/3

Since the width cannot be negative, we reject x = -9/3 and take x = 8.The length of the carpet is 3 + 3x = 3 + 3(8) = 27.

Therefore, the dimensions of the carpet are 8 feet by 27 feet.