Question

Difficulty: EasyQuadratic Equations and the Quadratic Formula

The length of a rectangle is 33 inches greater than its width. If the area of the rectangle is 1010 square inches, what is the width of the rectangle, in inches?

  1. A
    11
  2. 22Answer
  3. C
    3.53.5
  4. D
    55
  5. E
    77

Answer

The width of the rectangle is 22 inches.
The correct answer is the option stating that the width is 22 inches. This is obtained by expressing the length as w+3w+3 and setting the area equation to w(w+3)=10w(w+3) = 10. Expanding and setting it to zero gives the quadratic equation w2+3w10=0w^2 + 3w - 10 = 0, which factors to (w+5)(w2)=0(w+5)(w-2) = 0. Since width must be positive, the only valid solution is 22.

Step-by-Step Solution

1
Define variables for the dimensions of the rectangle based on the problem statement.
Let the width of the rectangle be ww inches. The length of the rectangle is w+3w + 3 inches.
The problem states that the length is 33 inches greater than the width.
2
Set up an equation representing the area of the rectangle.
The area is the product of width and length: w(w+3)=10w(w + 3) = 10.
The area of a rectangle is calculated as width times length, and the given area is 1010 square inches.
3
Rewrite the equation in standard quadratic form aw2+bw+c=0aw^2 + bw + c = 0.
Distribute ww to get w2+3w=10w^2 + 3w = 10, then subtract 1010 from both sides: w2+3w10=0w^2 + 3w - 10 = 0.
Standard quadratic form is required to solve the equation by factoring or using the quadratic formula.
4
Factor the quadratic equation.
(w+5)(w2)=0(w + 5)(w - 2) = 0.
Finding factors that multiply to 10-10 and add to 33 allows us to find the roots.
5
Solve for ww and apply real-world constraints.
w=5w = -5 or w=2w = 2. Since a physical width must be positive, discard 5-5, leaving w=2w = 2.
A dimension of a geometric shape cannot be negative.

Key Concept

Solving quadratic equations derived from geometric word problems by factoring or using the quadratic formula.
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