Turning real situations — area, growth, motion — into quadratics, and a bridge toward inequalities.
Quadratics are not a puzzle invented to torture you on a Friday. They are the natural language for any situation where a quantity gets multiplied by itself: the area of a plot of land (length times width), money that earns interest on its own interest, two moving things meeting somewhere down the road. Whenever an unknown shows up twice and the two copies get multiplied together, a square appears — and a square is a quadratic. This lesson is about hearing that pattern in plain English, turning the words into an equation, solving it, and then doing the one step beginners always skip: throwing out the answers the situation forbids.
By the end you'll have a reliable five-step recipe for modeling, you'll handle garden plots, growth-on-growth, and consecutive-number puzzles, and you'll be standing at the door of the next stage, where the sharp word "equals" softens into "at least" and "no more than." Throughout, the unknown and its solutions are violet, the standard-form coefficients keep their usual a-blue, b-teal, and c-amber, and any root we have to reject is marked in red.
Here is the secret that makes word problems stop being scary: you don't have to be clever, you have to be orderly. The same five moves work every time, whether the problem is about a garden, a bank balance, or two trains. Read them once, then watch them do all the heavy lifting in every example below.
Solving the quadratic and stopping. You found x = 4 or x = −6, you circle both, you move on — and you're wrong, because a width of −6 metres is nonsense. Step 5 is not optional polish; it is half the problem. A quadratic almost always gives two roots, and the situation usually wants one.
Area is the most natural home for a quadratic, because area is a product: length times width. If both the length and the width are built out of the same unknown, multiplying them squares that unknown. Let's do the garden from the picture, slowly, with the recipe.
The garden. A rectangular garden is 2 metres longer than it is wide, and its area is 24 m². How wide is it?
Step 1 — Name. Let the width be w metres. Step 2 — Relationship. The length is "2 more than the width," so the length is w + 2, and the area (length × width) equals 24. Step 3 — Write:
w(w + 2) = 24, which expands to w2 + 2w − 24 = 0.
Step 4 — Solve. We need two numbers that multiply to −24 and add to +2. Those are +6 and −4, so the equation factors as (w + 6)(w − 4) = 0, giving w = −6 or w = 4. Step 5 — Check. A width of −6 m is impossible, so we reject it. The width is 4 m and the length is 6 m. Quick sanity check: 4 × 6 = 24. ✓
A border variant. The same move covers paths and frames. If a square lawn of side s is wrapped by a 1 m walkway all the way around, the whole region is a bigger square of side s + 2 (one metre added on each of the two sides), so its area is (s + 2)2. Set that equal to whatever total area you're given and you again have a quadratic — and again you reject the negative root, because a side length can't be negative.
Slide the width and watch the rectangle and its area change. Stop when the area hits the dashed target of 24.
Here is the second place squares are born: when the same rate of change happens twice in a row. The trick is that the second change acts on a starting amount that the first change already grew — growth piles on top of growth, and that "on top of" is exactly multiplication. So two rounds of growing by rate x carry a factor of (1 + x)2. This is the same idea as interest earning interest.
Worked example. A workshop's output rose from 100 units to 144 units over two years, growing by the same rate each year. Find the yearly rate.
Name: let the yearly rate be x (as a decimal). Relationship & write: after one year output is 100(1 + x); after the second year that grows again by the same factor, giving 100(1 + x)2 = 144. Solve: divide by 100 to get (1 + x)2 = 1.44, take the square root: 1 + x = ±1.2. Check: the plus gives x = 0.2 = 20%; the minus gives 1 + x = −1.2, so x = −2.2, a rate of −220%, which is impossible — you can't lose more than everything. The rate is 20% per year.
If a quantity grows by rate x each round, multiply by (1 + x). If it shrinks by rate x, multiply by (1 − x). Two rounds means squaring that factor. A price that falls twice from 80 to 64.80 obeys 80(1 − x)2 = 64.80 → (1 − x)2 = 0.81 → 1 − x = 0.9 → x = 0.1 = 10%.
Set the yearly rate and watch 100 grow twice. Find the rate that lands exactly on the target of 144.
Strip away the scenery and every quadratic word problem has the same skeleton: somewhere a product of two unknown quantities appears, and since both quantities are built from your one unknown, the equation climbs to second degree. Number puzzles make this skeleton easy to see.
Consecutive integers. Two consecutive positive integers have a product of 56. Find them.
Name: let the smaller be n; the next integer is n + 1. Write: n(n + 1) = 56, i.e. n2 + n − 56 = 0. Solve: two numbers multiplying to −56 and adding to +1 are +8 and −7, so (n − 7)(n + 8) = 0 and n = 7 or n = −8. Check: we were told "positive," so reject −8. The integers are 7 and 8, and indeed 7 × 8 = 56. ✓
The same move handles two-digit-number puzzles (the value is 10·tens + units, and some product of the digits is given) and travel problems where two objects meet or one overtakes another — name the unknown time or distance, a product appears, solve, and check that a time isn't negative. The story changes; the skeleton doesn't.
Reveal the five modeling steps one at a time for the consecutive-integers problem, then reset.
Every equation in this whole stage asks one kind of question: what makes the two sides exactly equal? The answers are a handful of sharp points — for x2 = 9, exactly the two points x = 3 and x = −3, nothing in between.
But real life rarely asks for exactly. A bridge must hold at least a certain weight; a budget must stay no more than a certain amount; a signal must keep its squared strength under some limit. The moment you replace the "=" with ">", "<", "≥", or "≤", something beautiful happens: the few isolated points of an equation spread out into a whole stretch of answers. Ask x2 ≤ 9 — "the square is no more than 9" — and the answer is the entire interval −3 ≤ x ≤ 3, every number from −3 to 3. The two boundary points stay the same; now the space between them joins in.
Pick a relation and see how the solution set of "x² ? 9" changes from points to a whole stretch.
A quadratic equation pins down a place; a quadratic inequality describes a region. Same parabola, same boundary numbers — but instead of "where does the curve touch this height?" you ask "where is the curve below (or above) this height?" That single shift opens all of Stage 12.
A square shows up whenever your one unknown gets multiplied by another quantity made of that same unknown — an area (length × width), growth on top of growth ((1 ± x)²), or a product of consecutive numbers. The way through is always the same five steps: name the unknown, find the hidden equality, write the equation, solve it, and — the step that separates a right answer from a wrong one — check each root against reality and reject the impossible one. And when "exactly equal" relaxes into "at least" or "no more than," the lone points of a solution swell into an interval, which is exactly where we go next.
We open Stage 12, Inequalities, by treating ">" and "≤" as first-class citizens. You'll learn to solve quadratic inequalities like x2 − x − 6 > 0 by reading the sign of the parabola — turning today's single solution points into the boundaries of whole solution intervals.
Work each one out first, then open the answer to check your thinking.
Tap the answer you think is right.
This lesson targets Common Core A-CED.A.1 (create equations and inequalities from real situations), A-REI.B.4b (solve quadratics by inspection, factoring, and the quadratic formula, recognizing when a root must be rejected), and modeling practice MP4. The single non-negotiable habit to reinforce is Step 5: every root must be checked against the context, and impossible roots (negative lengths, sub−100% rates, fractional counts) discarded — this is the most common place strong students still lose points. The growth model uses (1 ± x)² rather than abstract compounding so the "growth on top of growth" structure is visible. The closing bridge to inequalities (Stage 12) is intentional: it previews how a single equation's isolated solution points widen into a whole solution interval once "=" becomes "≤" or "≥," which prevents the common misconception that an inequality "has one answer."