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Difficulty: MediumQuadratic Equations and the Quadratic Formula

A rectangular painting is 22 feet wide and 66 feet long. A wooden frame of uniform width xx feet is placed around the painting. If the total area of the painting and the frame is 2121 square feet, what is the width of the frame, in feet?

  1. A
    1.5
  2. B
    1.0
  3. 0.5Answer
  4. D
    4.5
  5. E
    8.0

Answer

The correct width of the frame is 0.5 feet.
The correct answer is 0.5 feet. To find the width of the frame, we define the total dimensions of the framed painting as (2+2x)(2 + 2x) and (6+2x)(6 + 2x). Setting their product equal to the total area of 21 square feet gives the quadratic equation 4x2+16x−9=04x^2 + 16x - 9 = 0. Solving this using the quadratic formula yields x=0.5x = 0.5 and x=−4.5x = -4.5. Since a physical width must be positive, the width of the frame is 0.5 feet.

Step-by-Step Solution

1
Set up the equation for the total area. The painting's dimensions are 22 feet by 66 feet. Adding a frame of uniform width xx on all sides increases both the width and the length by 2x2x.
The total dimensions are (2+2x)(2 + 2x) and (6+2x)(6 + 2x), and the total area is given by the equation: (2+2x)(6+2x)=21(2 + 2x)(6 + 2x) = 21.
To find the width of the frame, we must relate the final total area to the dimensions of the painting plus the frame.
2
Expand the equation and write it in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
12+4x+12x+4x2=21  ⟹  4x2+16x+12=21  ⟹  4x2+16x−9=012 + 4x + 12x + 4x^2 = 21 \implies 4x^2 + 16x + 12 = 21 \implies 4x^2 + 16x - 9 = 0.
Standard form is required to apply the quadratic formula or to factor the quadratic expression.
3
Solve the quadratic equation 4x2+16x−9=04x^2 + 16x - 9 = 0 using the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
x=−16±162−4(4)(−9)2(4)=−16±256+1448=−16±4008=−16±208x = \frac{-16 \pm \sqrt{16^2 - 4(4)(-9)}}{2(4)} = \frac{-16 \pm \sqrt{256 + 144}}{8} = \frac{-16 \pm \sqrt{400}}{8} = \frac{-16 \pm 20}{8}.
This formula provides the solutions to any quadratic equation.
4
Calculate the two possible values for xx and select the physically meaningful one.
x=−16+208=0.5x = \frac{-16 + 20}{8} = 0.5 or x=−16−208=−4.5x = \frac{-16 - 20}{8} = -4.5. Since width must be positive, the only valid solution is x=0.5x = 0.5.
A physical measurement like width cannot be negative.

Key Concept

Quadratic Equations and the Quadratic Formula
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Topics

Quadratic Equations and the Quadratic Formula

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