The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 20 (Intermediate Algebra) Quadratic Equation Applications

Overview

This lecture applies the quadratic-equation techniques from earlier in the course to real-world word problems. No new algebra is introduced; instead, the focus is on translating a verbal situation into a quadratic equation, solving it (by factoring or the quadratic formula), and then deciding which root is physically meaningful. The problem types covered are geometric area problems, work-rate ("teamwork") problems, distance–rate–time problems, and systems of two equations that reduce to a single quadratic. The lecture closes with a brief introduction to graphing a parabola by plotting points.

Key Concepts

Setting up a quadratic word problem. Translate the words into an equation, expand, and collect every term on one side so the equation has the form $ax^2 + bx + c = 0$ . Then either factor or apply the quadratic formula.

The quadratic formula. For $ax^2 + bx + c = 0$ , $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ Factoring is faster when the constant factors nicely into two integers whose product is $ac$ and whose sum is $b$ ; otherwise the formula always works.

Discarding unphysical roots. A quadratic gives two roots, but lengths, widths, speeds, times, and prices cannot be negative. Reject any root that produces a negative dimension and keep the positive one. Always sanity-check the surviving root against the original numbers.

Area problems.

Rectangle: $\text{area} = \text{length} \times \text{width}$ . If one side is described in terms of the other (e.g. "width is 3 more than length"), substitute to get one variable.
Square scaling: enlarging each side of a square of side $x$ by $6$ gives a new side $x+6$ and new area $(x+6)^2$ .
Framing problems: a picture of width $w$ and height $h$ surrounded by a frame of uniform width $x$ has outer dimensions $(w + 2x)$ by $(h + 2x)$ , because the frame adds $x$ on both sides.

Simultaneous (systems reducing to a quadratic). Given two equations in two unknowns, solve one for a variable and substitute into the other. Clearing a fraction often introduces an $x^2$ term, producing a quadratic.

Work-rate ("teamwork") problems. Rates add. If one worker finishes a job in $a$ hours and another in $b$ hours, their combined time $t$ satisfies $\frac{1}{a} + \frac{1}{b} = \frac{1}{t}.$ The intuition: a worker who finishes in $a$ hours completes $\tfrac{1}{a}$ of the job per hour, and the per-hour rates of all workers add to the combined per-hour rate. When something works against the job (e.g. a draining sink), subtract its rate. Clear denominators by multiplying through by the least common multiple of the denominators.

Distance–rate–time problems. Use $\text{rate} \times \text{time} = \text{distance}$ . Set up a relation for each leg of the trip; substituting one equation into the other typically yields a quadratic in the rate.

Graphing a parabola. A quadratic $y = ax^2 + bx + c$ graphs as a parabola. Build a table of values by substituting several $x$ -values and plotting the resulting points. The $y$ -intercept is found by setting $x = 0$ .

Worked Example

A picture measures $8 \text{ in} \times 12 \text{ in}$ and is surrounded by a frame of uniform width $x$ . The area of the frame equals the area of the picture. Find the width of the frame.

The picture's area is $8 \times 12 = 96$ , so the frame also covers $96$ , making the total outer area $192$ . The outer dimensions are $(8 + 2x)$ by $(12 + 2x)$ :

\begin{aligned} (8 + 2x)(12 + 2x) &= 192 \\ 4(4 + x)(6 + x) &= 192 \\ (4 + x)(6 + x) &= 48 \\ x^2 + 10x + 24 &= 48 \\ x^2 + 10x - 24 &= 0 \\ (x + 12)(x - 2) &= 0 \\ x &= -12 \quad \text{or} \quad x = 2. \end{aligned}

A frame width cannot be negative, so $x = 2$ in. Check: the outer rectangle is $12 \times 16 = 192$ , which is exactly twice the picture's $96$ , confirming the frame area equals the picture area.

Summary

Word problems become quadratics: translate to an equation, expand, set equal to zero, then factor or use the quadratic formula.
A quadratic yields two roots; discard any that give a negative length, time, speed, or price, and verify the survivor against the original data.
Rectangle area uses length × width; a uniform frame of width $x$ adds $2x$ to each outer dimension.
Systems of two equations are solved by isolating one variable and substituting, which often produces a quadratic.
Work-rate problems add rates: $\tfrac{1}{a} + \tfrac{1}{b} = \tfrac{1}{t}$ , subtracting any rate that works against the job; clear fractions by multiplying through by the LCM.
Distance problems use rate × time = distance, with each leg of the trip giving an equation; a parabola $y = ax^2 + bx + c$ is graphed by plotting a table of points.