Homework 13 quadratic equation word problems connect algebra to geometry, number patterns, and motion. Each scenario produces a quadratic equation whose roots (and sometimes vertex) carry the meaning of a length, an integer, or a time value.
Core translation pattern
Variable meaning: one unknown quantity is represented by a single variable, with units attached (feet, meters, seconds, integers).
Quadratic form: products like length × width, or motion models like height = −(constant)·t² + (constant)·t + (constant), generate a degree-2 equation.
Context filtering: negative lengths, negative times after launch, and other nonphysical values are excluded by constraints stated in the scenario.
Visualization: roots and vertex in a projectile model
Assignment set (modeled and solved)
Geometry and area
A rectangle and a square have the same area. The side length of the square is 20 ft. The rectangle’s length is 20 ft more than twice its width. The rectangle’s width and length are requested.
Model equation
The square’s area equals \(20^2 = 400\ \text{ft}^2\). The rectangle’s width is represented by \(w\) (ft), and the rectangle’s length is \(2w + 20\) (ft).
\[ w(2w + 20) = 400 \]
\[ 2w^2 + 20w - 400 = 0 \quad \Longrightarrow \quad w^2 + 10w - 200 = 0 \]
Solution and interpretation
\[ w = \frac{-10 \pm \sqrt{10^2 - 4(1)(-200)}}{2} = \frac{-10 \pm \sqrt{900}}{2} = \frac{-10 \pm 30}{2} \]
The two algebraic values are \(w = 10\) and \(w = -20\). A negative width is excluded by context, so the rectangle’s width is \(10\ \text{ft}\) and its length is \(2(10)+20 = 40\ \text{ft}\).
Consecutive odd integers
Three consecutive odd integers are represented by \(n\), \(n+2\), and \(n+4\). The product of the first and third integers equals twice the sum of all three integers plus 3.
Model equation
\[ n(n+4) = 2\big(n + (n+2) + (n+4)\big) + 3 \]
\[ n^2 + 4n = 2(3n+6) + 3 = 6n + 15 \quad \Longrightarrow \quad n^2 - 2n - 15 = 0 \]
Solution and interpretation
\[ n^2 - 2n - 15 = (n-5)(n+3) = 0 \quad \Longrightarrow \quad n = 5 \ \text{or}\ n = -3 \]
The two valid triples are \((5,7,9)\) and \((-3,-1,1)\). Both satisfy the condition and preserve the “odd integer” constraint.
Projectile motion (time of flight and maximum height)
A ball is launched straight upward from ground level with initial velocity 80 ft/s. A standard algebra-friendly model in feet uses \(s(t) = -16t^2 + 80t\), where \(s\) is height (ft) after \(t\) seconds.
Ground contact and roots
Ground level corresponds to \(s(t)=0\).
\[ -16t^2 + 80t = 0 \quad \Longrightarrow \quad -16t(t-5) = 0 \]
The roots are \(t=0\) and \(t=5\). The physically meaningful time of flight after launch is \(5\ \text{s}\).
Maximum height and vertex
For \(s(t) = at^2 + bt + c\) with \(a<0\), the maximum occurs at the vertex time \(t_v = -\dfrac{b}{2a}\).
\[ t_v = -\frac{80}{2(-16)} = 2.5\ \text{s} \]
\[ s(2.5) = -16(2.5)^2 + 80(2.5) = -16(6.25) + 200 = -100 + 200 = 100\ \text{ft} \]
The maximum height is \(100\ \text{ft}\), reached at \(2.5\ \text{s}\).
Summary table
| Scenario | Unknown | Quadratic equation | Meaningful result |
|---|---|---|---|
| Equal areas (square vs rectangle) | \(w\) = rectangle width (ft) | \(w(2w+20)=400\) | \(w=10\ \text{ft}\), length \(=40\ \text{ft}\) |
| Three consecutive odd integers | \(n\) = first odd integer | \(n(n+4)=2(3n+6)+3\) | \((5,7,9)\) or \((-3,-1,1)\) |
| Vertical motion (feet model) | \(t\) = time (s) | \(-16t^2+80t=0\) | Flight time \(5\ \text{s}\); max height \(100\ \text{ft}\) at \(2.5\ \text{s}\) |
Common pitfalls
Rejected roots: negative lengths and negative “time after launch” values are excluded even when algebra produces them.
Units: mixing feet and meters changes the gravity constant (commonly 16 in feet models and 4.9 in metric models).
Meaning of the vertex: maximum or minimum context depends on the sign of the leading coefficient \(a\).