Quadratic Formula Word Problems: Projectiles, Break-Even Prices and the Discriminant

In 30 seconds

Many word problems turn into an equation of the form $ ax^2 + bx + c = 0 $. The quadratic formula solves every one of them, even when the numbers will not factorise.
The method is always the same: write the model, set it equal to the target value, move everything to one side, read off $ a $, $ b $ and $ c $, then substitute.
The discriminant $ b^2-4ac $ answers a yes-or-no question before you solve: can this height, price or level ever be reached at all?
Most word problems give two roots, but the context often rules one out (negative time, a negative length). Say why you reject it.
Exam point: give answers to a sensible accuracy, with units, and finish with a sentence in the language of the question.

The case: a water rocket and a cupcake stall

A school science fair has two stands side by side. At the first, a team launches a water rocket straight up from a platform 1.5 m above the ground. It leaves the platform at 19.6 m/s. Ignoring air resistance, its height in metres after $ t $ seconds is

\[ h = 1.5 + 19.6t-4.9t^2 \]

A banner is strung 15 m above the field, and the judges want to know three things. How long does the rocket spend above the banner? When does it land? And could it ever reach the 25 m rooftop marker?

At the second stand, the same school’s bakery club sells cupcakes. From past fairs they expect to sell $ q = 300-40p $ cupcakes in a day at a price of $ p $ dollars. Each cupcake costs $1.50 to make, and renting the stand costs $150. Which prices let them break even, and which price gives the most profit?

These numbers are illustrative, chosen to keep the arithmetic clean, but both questions have exactly the shape you meet in exams, and both end in a quadratic that does not factorise nicely.

The concept: turning a question into “something squared equals zero”

A quadratic model has a curved graph with a single turning point: a peak for the rocket, a peak for the profit. Word problems almost always ask one of three things about such a curve:

When or where does it reach a given level? Set the model equal to that level and solve.
Can it reach that level at all? Check whether the resulting equation has real roots.
What is the highest or lowest value? Find the turning point, which sits exactly halfway between the roots.

The first two need the same move: rearrange a statement such as “the height is 15 m” until one side is zero. The curve meets the target level at the roots of that new quadratic. With decimals like 19.6 and 4.9 you can rarely spot factors, and that is what the formula is for.

Key termDiscriminant: the quantity $ \Delta = b^2-4ac $ inside the square root of the quadratic formula. If $ \Delta > 0 $ there are two real roots (the level is crossed twice), if $ \Delta = 0 $ there is one repeated root (the level is just touched at the turning point), and if $ \Delta < 0 $ there are no real roots (the level is never reached).

The formula

For any equation $ ax^2 + bx + c = 0 $ with $ a \ne 0 $, the roots are^[1]

\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

Here $ x $ is the unknown (time, price, length), $ a $ is the coefficient of $ x^2 $, $ b $ is the coefficient of $ x $, and $ c $ is the constant term, all read after the equation has been rearranged so the right-hand side is zero. The $ \pm $ gives the two roots, and $ b^2-4ac $ is the discriminant. The line of symmetry of the graph, where the peak or trough sits, is at $ x = -\dfrac{b}{2a} $, which is the average of the two roots^[5].

Chapter 2 of Mathematics Made Simple builds the quadratic formula from completing the square, then puts it to work.Get the textbook · $12

Why this formula, and when not to use it

The quadratic formula is completing the square done once, in general, which is why it never fails. It is the right tool here because both cases produce coefficients such as 4.9 and 13.5, and factorising those would be guesswork.

It is not always the best tool. If the equation factorises easily, such as $ t^2-5t + 6 = 0 $, factorising is quicker and less error-prone. If the question only asks for the maximum height or profit, you need the vertex, not the roots, so use $ x = -b/2a $ or completing the square. And if the model is not really quadratic, the formula gives exact answers to the wrong question. The projectile model $ h = h_0 + ut-\tfrac{1}{2}gt^2 $ with $ g = 9.8 $ m/s² is the standard constant-acceleration result from introductory physics^[4], and it is only as good as its assumption that gravity is the only force that matters.

Setting it up

Every word problem in this post follows the same four steps.

Name the variable and its units. For the rocket, $ t $ is time in seconds after launch, so only $ t \ge 0 $ makes sense. For the stall, $ p $ is the price in dollars, and since sales $ q = 300-40p $ cannot be negative, $ 0 \le p \le 7.5 $.
Write the condition as an equation. “Above the banner” becomes $ h(t) = 15 $ at its edges. “Break even” becomes profit $ = 0 $.
Rearrange to $ ax^2 + bx + c = 0 $, ideally with $ a $ positive.
Read off $ a $, $ b $, $ c $ with their signs, check the discriminant, then substitute.

For the cupcake stall, the profit needs building first. Each cupcake earns $ p-1.50 $ dollars over its cost, and $ 300-40p $ are sold, so before the stand rent the margin is $ (p-1.5)(300-40p) $. Expanding gives $ 300p-40p^2-450 + 60p = -40p^2 + 360p-450 $. Subtract the $150 rent to get the daily profit $ \pi = -40p^2 + 360p-600 $, which can also be written

\[ \pi = -40(p^2-9p + 15) \]

Solving it

Worked example

The rocket has $ h(t) = 1.5 + 19.6t-4.9t^2 $. Find (a) the times when it passes the 15 m banner, and how long it spends above it, and (b) the time when it lands.

Step 1(a) Set $ h = 15 $: $ 1.5 + 19.6t-4.9t^2 = 15 $. Move everything to the right to make $ a $ positive: $ 4.9t^2-19.6t + 13.5 = 0 $. So $ a = 4.9 $, $ b = -19.6 $, $ c = 13.5 $.

Step 2Discriminant: $ b^2-4ac = (-19.6)^2-4(4.9)(13.5) $ $ = 384.16-264.6 $ $ = 119.56 $. It is positive, so the rocket crosses 15 m twice.

Step 3$ t = \dfrac{19.6 \pm \sqrt{119.56}}{9.8} = \dfrac{19.6 \pm 10.934}{9.8} $

Step 4$ t = \dfrac{8.666}{9.8} \approx 0.88 $ s (going up) and $ t = \dfrac{30.534}{9.8} \approx 3.12 $ s (coming down). Time above the banner: $ 3.116-0.884 \approx 2.23 $ s.

Step 5(b) Landing means $ h = 0 $: $ 4.9t^2-19.6t-1.5 = 0 $, with $ a = 4.9 $, $ b = -19.6 $, $ c = -1.5 $.

Step 6$ b^2-4ac = 384.16 + 29.4 = 413.56 $, and $ \sqrt{413.56} \approx 20.336 $.

Step 7$ t = \dfrac{19.6 \pm 20.336}{9.8} $, giving $ t \approx 4.08 $ s or $ t \approx -0.075 $ s. Reject the negative root: the rocket did not exist in flight before launch.

The rocket spends about 2.23 s above the banner and lands about 4.08 s after launch. Notice that the two banner times, 0.88 s and 3.12 s, are equally spaced either side of $ t = 2 $. That is the line of symmetry $ t = -b/2a = 19.6/9.8 = 2 $, where the rocket peaks.

The third question, “could it reach 25 m?”, needs no solving at all. Setting $ h = 25 $ gives $ 4.9t^2-19.6t + 23.5 = 0 $, whose discriminant is $ 384.16-4(4.9)(23.5) = 384.16-460.6 = -76.44 $. It is negative, so there is no real time at which the height is 25 m. The peak height confirms it: $ h(2) = 1.5 + 39.2-19.6 = 21.1 $ m.

Height of the water rocket, h = 1.5 + 19.6t − 4.9t² metres

h(t), the rocket's height

Figure 1. The two roots of h = 15 bracket the 2.23 seconds the rocket spends above the banner line (shaded). The line h = 25 sits above the peak, so the equation h = 25 has no real roots.

Worked example

The bakery club’s profit is $ \pi(p) = -40p^2 + 360p-600 $ dollars. Find the break-even prices and the price that maximises profit.

Step 1Break even: $ -40p^2 + 360p-600 = 0 $. Divide every term by $ -40 $ to get smaller numbers and a positive $ a $: $ p^2-9p + 15 = 0 $. So $ a = 1 $, $ b = -9 $, $ c = 15 $.

Step 2Discriminant: $ (-9)^2-4(1)(15) = 81-60 = 21 $. Positive, so there are two break-even prices.

Step 3$ p = \dfrac{9 \pm \sqrt{21}}{2} = \dfrac{9 \pm 4.583}{2} $

Step 4$ p \approx \dfrac{4.417}{2} \approx 2.21 $ or $ p \approx \dfrac{13.583}{2} \approx 6.79 $. Both lie inside $ 0 \le p \le 7.5 $, so both are valid.

Step 5Best price: halfway between the roots, $ p = -\dfrac{b}{2a} = \dfrac{360}{80} = 4.50 $.

Step 6Maximum profit: $ \pi(4.5) = -40(20.25) + 360(4.5)-600 $ $ = -810 + 1620-600 $ $ = 210 $.

Any price strictly between about $2.21 and $6.79 makes a profit. The best price is $4.50, at which the club sells $ 300-40(4.5) = 120 $ cupcakes and makes $210 for the day.

Now the discriminant earns its keep a second time. Suppose the stand rent is $ F $ dollars instead of $150. Then $ \pi = -40p^2 + 360p-450-F $, and break-even means $ 40p^2-360p + (450 + F) = 0 $. The discriminant is $ 360^2-4(40)(450 + F) $ $ = 129\,600-72\,000-160F $ $ = 57\,600-160F $. A break-even price exists only if $ 57\,600-160F \ge 0 $, that is $ F \le 360 $. With a $400 rent, no price at all can cover costs, which is the dashed curve in Figure 2.

Worked example

(Exam style.) The rocket team wants to reach the 25 m marker by changing the launch speed $ u $ m/s, still from the 1.5 m platform, so $ h = 1.5 + ut-4.9t^2 $. Find the smallest launch speed that works, to 3 significant figures.

Step 1Set $ h = 25 $ and rearrange: $ 4.9t^2-ut + 23.5 = 0 $, with $ a = 4.9 $, $ b = -u $, $ c = 23.5 $.

Step 2The rocket reaches 25 m if and only if this has a real root: $ b^2-4ac \ge 0 $.

Step 3$ u^2-4(4.9)(23.5) \ge 0 $, so $ u^2 \ge 460.6 $.

Step 4Speed is positive, so $ u \ge \sqrt{460.6} \approx 21.46 $. The minimum launch speed is 21.5 m/s (3 s.f.).

Step 5At exactly this speed $ \Delta = 0 $ and the single root is $ t = \dfrac{u}{2(4.9)} = \dfrac{21.46}{9.8} \approx 2.19 $ s: the rocket just touches 25 m at its peak.

Zero discriminant is the boundary between “never gets there” and “passes through twice”.

The Practice Questions book drills quadratic word problems, discriminant conditions and more, all fully worked.Get 1,565 questions · $9

What the answers tell us

Do the rocket answers make sense? A rocket leaving at 19.6 m/s slows by 9.8 m/s every second, so it should stop rising after exactly 2 s, and it does. It falls past the platform to the ground, so the flight lasts 4.08 s rather than 4 s: the extra 0.08 s is the 1.5 m platform. Units check out too: metres in, seconds out.

How sensitive is the time above the banner? The gap between the two roots is $ \dfrac{2\sqrt{\Delta}}{2a} = \dfrac{\sqrt{\Delta}}{a} $, here with $ a = 4.9 $. Raise the banner and $ \Delta $ shrinks, until at 21.1 m (the peak) the time above is zero. It is also an exam shortcut: $ \sqrt{\Delta}/|a| $ gives “how long above” without finding either time.

What does the cupcake answer mean for the club? The profit curve is a $ \cap $, so profit is positive between the roots. The band of profitable prices is $ \sqrt{21} \approx \$4.58 $ wide, centred on $4.50. Higher costs push the curve down and narrow the band; at a $360 rent it shrinks to the single price $4.50. All of this rests on the demand line $ q = 300-40p $ holding across that range, an assumption about the market rather than a fact of algebra.

Daily profit against cupcake price $ per day

fixed cost $150: π = −40p² + 360p − 600fixed cost $400: never breaks even

Figure 2. With $150 of fixed costs the profit curve cuts the axis twice, so there is a band of profitable prices (shaded). Raise fixed costs to $400 and the whole curve drops below zero: the discriminant of the break-even equation turns negative.

If you want the whole chapter on quadratics, from factorising to the discriminant and inequalities, it is in the Mathematics Made Simple textbook, which is mapped to AP Calculus, Cambridge 9709 and IB Mathematics: analysis and approaches.

Common mistakeReading $ a $, $ b $ and $ c $ before rearranging. From $ 1.5 + 19.6t-4.9t^2 = 15 $, some students take $ c = 1.5 $ and get the wrong roots. The constant term is only $ c $ once the right-hand side is zero: here $ c = 13.5 $ (or $ -13.5 $ if you keep $ a $ negative). Always write the rearranged equation out in full first.

Common mistakeLosing the sign of $ b $. With $ b = -19.6 $, the formula starts $ -b = +19.6 $, and $ b^2 = (-19.6)^2 = +384.16 $. Put negative values in brackets on your calculator, or $ -19.6^2 $ will come out as $ -384.16 $.

Exam tipThe quadratic formula and the discriminant are in Cambridge 9709 Paper 1 (topic 1.1, Quadratics), where questions on “the set of values of $ k $” for which an equation has real or no roots come up often^[2]. IB Mathematics: analysis and approaches lists the quadratic formula and the discriminant in SL topic 2.7^[3]. For AP Calculus they are assumed prerequisite algebra. In every syllabus, state why you reject a root (“$ t \ge 0 $”) rather than just crossing it out.

Get both Mathematics Made Simple books for $16: the full quadratics chapter plus hundreds more questions to master it.Both books · $16

Practice questions

Try each one on paper first, then open the worked answer. A calculator is allowed. You will find many more like these, each with a fully worked answer, in the Mathematics Made Simple practice questions book.

EasyQ1. Solve $ 2x^2-7x-3 = 0 $, giving your answers to 3 significant figures. Show worked answer

$ a = 2 $, $ b = -7 $, $ c = -3 $.
$ b^2-4ac = 49 + 24 = 73 $.
$ x = \dfrac{7 \pm \sqrt{73}}{4} = \dfrac{7 \pm 8.544}{4} $.
$ x \approx 3.89 $ or $ x \approx -0.386 $.

EasyQ2. A rectangular lawn 12 m by 8 m is surrounded by a path of constant width $ x $ m. The lawn and path together cover 192 m². Find $ x $. Show worked answer

Outer rectangle: $ (12 + 2x)(8 + 2x) = 192 $.
Expand: $ 96 + 40x + 4x^2 = 192 $, so $ 4x^2 + 40x-96 = 0 $, or $ x^2 + 10x-24 = 0 $.
$ x = \dfrac{-10 \pm \sqrt{100 + 96}}{2} = \dfrac{-10 \pm 14}{2} $, so $ x = 2 $ or $ x = -12 $.
A width cannot be negative, so the path is 2 m wide.

MediumQ3. A cricket ball is hit upwards so that its height is $ h = 2 + 12t-5t^2 $ metres after $ t $ seconds. When does it hit the ground? Give your answer to 3 significant figures. Show worked answer

Set $ h = 0 $ and multiply by $ -1 $: $ 5t^2-12t-2 = 0 $.
$ b^2-4ac = 144 + 40 = 184 $, $ \sqrt{184} \approx 13.565 $.
$ t = \dfrac{12 \pm 13.565}{10} $, so $ t \approx 2.56 $ or $ t \approx -0.156 $.
Reject the negative time: the ball lands after about 2.56 s.

MediumQ4. A small print shop’s daily profit from $ q $ posters is $ \pi = -0.2q^2 + 16q-180 $ dollars. Find the range of whole numbers of posters for which the shop makes a profit. Show worked answer

Break even: $ -0.2q^2 + 16q-180 = 0 $. Multiply by $ -5 $: $ q^2-80q + 900 = 0 $.
$ b^2-4ac = 6400-3600 = 2800 $, $ \sqrt{2800} \approx 52.915 $.
$ q = \dfrac{80 \pm 52.915}{2} \approx 13.54 $ or $ 66.46 $.
Profit is positive between the roots, so for whole numbers $ 14 \le q \le 66 $.
Check: $ \pi(14) = 4.8 > 0 $ and $ \pi(13) = -5.8 < 0 $.

ExamQ5. A stone is thrown upwards from the top of a 20 m cliff. Its height above the sea after $ t $ seconds is $ h = 20 + 15t-4.9t^2 $ metres. (a) Find when the stone hits the sea, to 3 significant figures. [3] (b) Show that the stone never reaches 35 m above the sea. [2] (c) Find, to 3 significant figures, how long the stone is more than 25 m above the sea. [3] Show worked answer

(a) $ h = 0 $: $ 4.9t^2-15t-20 = 0 $ [1]. $ \Delta = 225 + 392 = 617 $ [1]. $ t = \dfrac{15 + \sqrt{617}}{9.8} = \dfrac{15 + 24.839}{9.8} \approx 4.07 $ s; the other root is negative and rejected [1].
(b) $ h = 35 $: $ 4.9t^2-15t + 15 = 0 $ [1]. $ \Delta = 225-294 = -69 < 0 $, so there are no real solutions and 35 m is never reached [1].
(c) $ h = 25 $: $ 4.9t^2-15t + 5 = 0 $, $ \Delta = 225-98 = 127 $ [1]. $ t = \dfrac{15 \pm \sqrt{127}}{9.8} \approx 0.381 $ or $ 2.68 $ [1]. Time above 25 m: $ 2.681-0.381 \approx 2.30 $ s [1]. (Shortcut: $ \sqrt{127}/4.9 \approx 2.30 $.)

Frequently asked questions

How do I know a word problem needs the quadratic formula?

Look for a squared term in the model: height under gravity, the area of a shape whose sides both change, or revenue when price affects sales. Rearrange to $ ax^2 + bx + c = 0 $, try factorising for a few seconds, and if that fails, use the formula.

What does a negative discriminant mean in a real problem?

It means the target is never reached: the ball never gets that high, or the business never breaks even. In exams, the instruction “show that” something cannot happen is often a hint to compute the discriminant and show it is negative.

Why do I get two answers, and which one is right?

The curve can meet a level twice, so the algebra offers two roots. Sometimes both are meaningful (the rocket passes 15 m going up and coming down; there is a low and a high break-even price). Sometimes the context rules one out, such as a negative time or length. Check each root against the domain you wrote down at the start, and say why you reject one.

References

Abramson, J. et al. (2021) College Algebra 2e, section 2.5 “Quadratic Equations”. Houston, TX: OpenStax. https://openstax.org/details/books/college-algebra-2e
Cambridge University Press & Assessment. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge International Education. cambridgeinternational.org
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021). Geneva: IBO.
Ling, S. J., Sanny, J. and Moebs, W. (2016) University Physics Volume 1, section 4.3 “Projectile Motion”. Houston, TX: OpenStax. https://openstax.org/details/books/university-physics-volume-1
Stewart, J., Redlin, L. and Watson, S. (2016) Precalculus: Mathematics for Calculus, 7th edn. Boston, MA: Cengage Learning.

Quadratic Formula Word Problems: Projectiles, Break-Even Prices and the Discriminant

The case: a water rocket and a cupcake stall

The concept: turning a question into “something squared equals zero”

The formula

Why this formula, and when not to use it

Setting it up

Solving it

What the answers tell us

Practice questions

Frequently asked questions

How do I know a word problem needs the quadratic formula?

What does a negative discriminant mean in a real problem?

Why do I get two answers, and which one is right?

References

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