Applications of Quadratics

Intuition

You can now solve any quadratic: factor it when a factorization is easy to spot, and reach for the quadratic formula otherwise. That skill is only worth having if you can point it at something. Plenty of ordinary questions hide a quadratic equation inside them. When does a thrown ball land, how wide is a garden of a given area, and what price brings in a target amount of money? The work splits into two halves. First you translate the words into an equation, naming the unknown and writing down what the situation says about it. Then you solve, and here comes the part that is new: solving usually hands you two real roots. It can also hand you one repeated root, or, as you will see near the end of this lesson, no real root at all, only a complex pair. When two real roots show up, watch carefully: often the situation allows only one of them, but sometimes, as later examples show, it allows both. A length cannot be negative, the flight’s clock cannot read a time before the throw, and you cannot have $-8$ chairs. Deciding which root the world keeps and which it throws away is the real content of this lesson.

Learning goals

Translate the words into a quadratic in standard form
Discard a root the situation forbids, and say why
Apply $h = -16t^2 + v_0 t + h_0$ to projectile height
Build a quadratic from an area or a right triangle
Read a negative discriminant as a height never reached

From words to a quadratic

Every problem in this lesson yields to the same four steps. They are worth stating plainly, because skipping the last two is the most common way applied problems go wrong.

Name the unknown. Choose a letter for the quantity you are asked to find, and write down, in terms of that letter, every other quantity the problem mentions. If the width is $w$ , then a length “three more than the width” is $w + 3$ , not a second free letter.
Translate to an equation. Turn the sentence that ties the quantities together (“the area is $40$ ”, “the height is $0$ ”, “the product is $56$ ”) into a single equation. Gather everything on one side so it reads $ax^2 + bx + c = 0$ , standard form.
Solve, then interpret. Solve the quadratic by whichever tool is convenient: factor when a factorization is easy to spot, otherwise use the quadratic formula. Then look at each real root and ask whether the situation permits it. Reject any root that is physically impossible, and say why: a negative length, a negative time, a count that is not a positive whole number.
Answer in a sentence, with units. A bare number is not an answer to a word problem. State the result in words, carry the correct units, and, when it helps, check it against the original sentence.

Step 3 is the heart of it. A quadratic can have zero, one, or two real roots. But the equation on its own cannot tell you which real root belongs to your problem; only the meaning of the letter can. In a great many of these problems the two roots come out with opposite signs, one positive and one negative, and the reason is not luck.

Projectile motion

Throw a ball straight up, or watch a rocket coast upward once its engines cut off. Either way, its height above the ground changes with time in a way physics describes with a quadratic. Near the surface of the Earth, an object moving under gravity alone, with nothing else pushing it, has height

h = -16t^2 + v_0\,t + h_0 \qquad (\text{height in feet, time } t \text{ in seconds}).

Each term earns its place. The starting height $h_0$ is where the object begins, at $t = 0$ . The term $v_0\,t$ is the height it would gain from its initial upward speed $v_0$ (in feet per second) if nothing pulled it back. The term $-16t^2$ is gravity pulling it down: a falling body drops about $16t^2$ feet in $t$ seconds, and the minus sign points that motion downward. In metric units the same pull of gravity is written with a smaller number. The model reads $h = -4.9t^2 + v_0\,t + h_0$ with the height in meters and $v_0$ in meters per second. Two questions cover almost everything you will be asked, and each is just a quadratic set equal to a known height.

To ask when the object hits the ground, set the height to $0$ and solve $h = 0$ . To ask when the object is at a particular height $H$ , set $h = H$ and solve. Either way you finish with the interpretation step, throwing out any negative time.

Worked example 1 When does a thrown ball hit the ground?

A ball is thrown upward so that its height in feet after $t$ seconds is $h = -16t^2 + 32t + 48$ . When does it hit the ground?

Hitting the ground means the height is $0$ , so set $h = 0$ :

-16t^2 + 32t + 48 = 0.

Every coefficient is divisible by $-16$ , and dividing by it clears the messy numbers. Dividing a right-hand side of $0$ by anything leaves $0$ :

t^2 - 2t - 3 = 0.

This factors, since $-3$ and $+1$ multiply to $-3$ and add to $-2$ :

(t - 3)(t + 1) = 0 \quad\Rightarrow\quad t = 3 \ \text{ or }\ t = -1.

Now interpret. The time $t = -1$ second would be one second before the throw, which is not part of the flight, so discard it. The ball hits the ground at $t = 3$ seconds.

Setting the height to a nonzero value is no harder, but the interpretation changes. An object on the way up and then on the way down can pass through the same height twice, so both positive roots can be real answers.

Worked example 2 When is the ball at a given height?

A ball launched from the ground has height $h = -16t^2 + 64t$ feet after $t$ seconds. At what times is it $48$ feet high?

Set the height equal to $48$ :

-16t^2 + 64t = 48.

Bring everything to one side and divide by $-16$ :

-16t^2 + 64t - 48 = 0 \quad\Rightarrow\quad t^2 - 4t + 3 = 0.

Factor, using $-1$ and $-3$ :

(t - 1)(t - 3) = 0 \quad\Rightarrow\quad t = 1 \ \text{ or }\ t = 3.

Both roots are positive, so this time both survive the interpretation step. The ball is $48$ feet high at $t = 1$ second, on the way up, and again at $t = 3$ seconds, on the way back down.

Check your understanding

The same ball has height $h = -16t^2 + 64t$ feet after $t$ seconds. At what times is it $40$ feet high? (This one does not factor over whole numbers.)

t = 2 \pm \dfrac{\sqrt{6}}{2}

t = 4 \pm \sqrt{6}

t = 2 \pm \sqrt{6}

t = 8 \pm 2\sqrt{6}

There is one more thing the height model can tell you: if the quadratic $h = H$ has a negative discriminant, the object never reaches height $H$ at all. Take the ball of Worked Example 2 again, with $h = -16t^2 + 64t$ , and ask whether it ever reaches $80$ feet:

\begin{gathered} -16t^2 + 64t = 80 \\ t^2 - 4t + 5 = 0 \end{gathered}

Its discriminant is

(-4)^2 - 4(1)(5) = -4.

Since $\sqrt{-4} = 2i$ , its roots are complex, and a complex time is no time at all. A negative discriminant here is the algebra’s way of reporting that the object simply never gets that high. Indeed, this ball’s greatest height, at $t = 2$ seconds, is only $64$ feet. Completing the square confirms it: $-16t^2 + 64t = -16(t - 2)^2 + 64$ , and since $(t - 2)^2$ is never negative, subtracting $16$ times it can never add height back above $64$ .

Check your understanding

The same ball has height $h = -16t^2 + 64t$ feet after $t$ seconds. Does it ever reach $100$ feet?

A Yes, at

t = 2

seconds B Yes, at two different times C No, the discriminant is negative D No, because the ball never leaves the ground

Check your understanding

A rock is dropped from a bridge $144$ feet above a river, so its height is $h = -16t^2 + 144$ . When does it reach the water?

3

seconds B

9

seconds C

12

seconds D

-3

seconds

Geometry: area and right triangles

Geometry is a rich source of quadratics, because area multiplies two lengths together, and if both lengths are written with the same unknown, that product is a quadratic. The reject-the-negative-root step is especially clean here, since a length is never negative.

A rectangle is $3$ centimeters longer than it is wide, and its area is $40$ square centimeters. Name the width $w$ , so the length is $w + 3$ , as in the picture.

A rectangle with width w and length w + 3. Both sides share the one unknown w, so its area w(w + 3) is a quadratic in w.

Worked example 3 A rectangle with area $40$

Find the rectangle’s dimensions. Area is length times width:

w(w + 3) = 40.

Expand and set to standard form:

w^2 + 3w - 40 = 0.

Factor, using $+8$ and $-5$ (they multiply to $-40$ and add to $+3$ ):

(w + 8)(w - 5) = 0 \quad\Rightarrow\quad w = -8 \ \text{ or }\ w = 5.

A width of $-8$ centimeters is impossible, so discard it. The width is $w = 5$ centimeters, and the length is $w + 3 = 8$ centimeters. As a check, $5 \times 8 = 40$ , the given area.

A right triangle brings in the Pythagorean theorem: the two legs and the hypotenuse satisfy $\text{leg}^2 + \text{leg}^2 = \text{hypotenuse}^2$ . A $3$ - $4$ - $5$ triangle checks it: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ . When the two legs are written with a shared unknown, squaring them produces a quadratic.

The longer leg of a right triangle is $7$ centimeters longer than the shorter leg, and the hypotenuse is $13$ centimeters. Let the shorter leg be $x$ , so the longer leg is $x + 7$ , as in the picture.

A right triangle with legs x and x + 7 and hypotenuse 13. The legs share the one unknown x, so the Pythagorean relation is a quadratic in x.

Worked example 4 A right triangle by the Pythagorean theorem

Find the two legs. The Pythagorean theorem ties them together:

x^2 + (x + 7)^2 = 13^2.

Expand $(x + 7)^2 = x^2 + 14x + 49$ and $13^2 = 169$ :

x^2 + x^2 + 14x + 49 = 169 \quad\Rightarrow\quad 2x^2 + 14x - 120 = 0.

Divide by $2$ , then factor:

x^2 + 7x - 60 = 0 \quad\Rightarrow\quad (x + 12)(x - 5) = 0 \quad\Rightarrow\quad x = -12 \ \text{ or }\ x = 5.

A leg of $-12$ centimeters is impossible, so discard it. The shorter leg is $x = 5$ centimeters and the longer leg is $x + 7 = 12$ centimeters. As a check, $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ .

A border of uniform width around a picture or a path around a pool follows the same pattern. To see it, notice that adding a strip of width $x$ to every side increases each outer dimension by $2x$ , one $x$ on each end.

Check your understanding

A photograph $5$ inches by $7$ inches is put in a frame of uniform width $x$ , so the outer rectangle measures $(5 + 2x)$ by $(7 + 2x)$ . If the outer area is $63$ square inches, how wide is the frame?

7

inches B

2

inches C

1

inch D

-7

inches

Number and money problems

Not every quadratic wears a picture. Some come from plain statements about numbers, and some from money, where a price times a quantity gives revenue. The method does not change; only the reason for rejecting a root does. For whole-number problems, a root that is negative or fractional gets thrown out because the answer must be a counting number.

Worked example 5 Consecutive integers with a given product

The product of two consecutive positive integers is $132$ . Find the integers.

Consecutive integers differ by $1$ , so name them $n$ and $n + 1$ . Their product is $132$ :

n(n + 1) = 132 \quad\Rightarrow\quad n^2 + n - 132 = 0.

Factor, using $+12$ and $-11$ :

(n + 12)(n - 11) = 0 \quad\Rightarrow\quad n = -12 \ \text{ or }\ n = 11.

The problem asks for positive integers, so discard $n = -12$ . Then $n = 11$ and $n + 1 = 12$ , and indeed $11 \times 12 = 132$ .

Money problems often set a revenue target. If raising the price sells fewer units, revenue (price times quantity) is a quadratic in the price. Then a target revenue is that quadratic set equal to a number. Here both roots can be positive and meaningful, two different prices bringing in the same money, so you read the question carefully to see which one it wants.

Worked example 6 A price that hits a target revenue

A vendor finds that at a price of $p$ dollars, the number of sandwiches sold in a day is $120 - 10p$ . The daily revenue is price times quantity. What price brings in $350$ dollars?

Write revenue as price times quantity and set it to the target of $350$ dollars:

p(120 - 10p) = 350 \quad\Rightarrow\quad 120p - 10p^2 = 350.

Gather to standard form and divide by $-10$ :

-10p^2 + 120p - 350 = 0 \quad\Rightarrow\quad p^2 - 12p + 35 = 0.

Factor, using $-5$ and $-7$ :

(p - 5)(p - 7) = 0 \quad\Rightarrow\quad p = 5 \ \text{ or }\ p = 7.

Both prices are positive, and each keeps the quantity $120 - 10p$ positive (selling $70$ at $5$ dollars, or $50$ at $7$ dollars). So both are genuine answers: a price of $5$ dollars and a price of $7$ dollars each bring in $350$ dollars. If the vendor wanted the lower price, it is $5$ dollars.

Check your understanding

Two positive numbers differ by $4$ , and their product is $45$ . What is the larger number?

5

13

-9

9

Common mistakes

Keeping a root the situation forbids. A width, a time, and a count cannot be negative. When a quadratic gives one positive and one negative root, the negative one is almost always rejected. When you reject it, say which reason applies (negative length, negative time, impossible count).
Discarding a root that is actually valid. The reject step is not “always throw away one root.” A ball can be at the same height twice, and a target revenue can occur at two prices. If both roots are positive and the situation allows both, keep both.
Introducing a second letter. If the length is “three more than the width”, write it as $w + 3$ , not as a new variable $\ell$ . Using one unknown is what makes the equation a single quadratic in one variable.
Forgetting that a border adds to both ends. A frame or path of width $x$ increases each outer dimension by $2x$ , not $x$ , because it wraps around both sides.
Answering with the wrong quantity. After solving for the width, a question that asks for the length or the area still needs one more step. Read what the problem actually asks, then convert.
Dropping the units, or the sentence. A word problem wants a stated answer with units. " $w = 5$ " is only half done; “the width is $5$ centimeters” is the answer.

Key takeaways

Model an applied problem in four steps. First name the unknown, then translate the words into a quadratic in standard form. Next solve it, factoring when a factorization is easy to spot and otherwise using the quadratic formula, then interpret the result and answer in a sentence with units.
Interpretation is the point of the lesson: keep only the real roots the situation allows, and say why you discard the rest. Often that keeps one of two roots, but sometimes it keeps both, and sometimes (a negative discriminant) no real root fits at all.
Projectile height near the Earth is $h = -16t^2 + v_0 t + h_0$ in feet (or $-4.9t^2 + v_0 t + h_0$ in meters). Set $h = 0$ for “hits the ground” and $h = H$ for “reaches height $H$ ”; a negative discriminant means the object never reaches that height.
Geometry gives quadratics through area: a rectangle whose sides share one unknown, or a border of width $x$ that adds $2x$ to each dimension. It also gives them through the Pythagorean theorem, when two legs share an unknown. Counts and money problems reduce to the same equation-and-interpret method.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Quadratic word problems (standard form) (Khan Academy)
Real world examples of quadratic equations (Math is Fun)
Quadratic word problems: projectile motion (Purplemath)
Solve applications modeled by quadratic equations (free textbook) (OpenStax)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why w(w + k) = A splits into a positive root and a negative root, for k > 0 and A > 0

Why w(w + k) = A splits into a positive root and a negative root, for k > 0 and A > 0#

Take the common shape of an area or product problem. Some positive quantity $w$ times a larger quantity $w + k$ (with $k > 0$ ) is set equal to a positive amount $A$ . Multiplying out and moving $A$ across gives the standard form

w^2 + kw - A = 0.

If $r$ and $s$ are the two roots, the quadratic factors as $(w - r)(w - s)$ , which expands to $w^2 - (r + s)w + rs$ . Matching that against $w^2 + kw - A$ term by term, the constant terms give $rs = -A$ : the two roots multiply to $-A$ , which is negative because $A > 0$ .

Its discriminant is $k^2 + 4A$ , which is positive, so both roots are real. A product of two real numbers is negative exactly when the two numbers have opposite signs. So one root is positive and the other is negative, with no other possibility. The quantity $w$ stands for a length or a count, which must be positive, so the negative root cannot be the answer and is discarded. For a length, that settles it: the positive root is exactly the answer. For a count, the positive root is the only candidate left, though you still confirm it comes out to a whole number, since not every equation of this shape happens to have one. Either way, at most one of the two roots can ever be the situation’s answer.

This is why an area or product problem of this form never leaves you guessing between two believable answers. The algebra produces two numbers, but the sign of one of them rules it out. Not every application has this shape, and a projectile can legitimately pass through a height twice. But when the leading coefficient is positive and the constant term is negative, the roots have opposite signs, so at most one survives.

A bit of history (optional)

The $-16t^2$ you leaned on all lesson is one of the hardest-won terms in science, and the hard part was the square.

For nearly two thousand years the accepted account of falling came from Aristotle. A heavier object falls faster, he held, in proportion to its weight. It is not a foolish idea. Drop a stone and a feather together and the stone wins easily. Air resistance holds the feather back far more than it holds back the stone, so the everyday observation does seem to confirm it.

In the early 1600s Galileo went after the question with a ramp. A ball dropped straight down moves far too fast to time by hand. So he rolled balls down a gently sloped board, slowing the motion until he could measure it, and then compared the distance traveled against the time.

What he found is the shape of this entire lesson. Distance was not proportional to time. It was proportional to the square of the time. A ball that covers one unit in the first second has covered a total of four units by two seconds, and nine units by three.

That is the whole source of the $t^2$ . The $16$ is only a matter of units, feet rather than meters, which is why the metric model reads $-4.9t^2$ instead. Without air pushing back, Galileo’s principle is that every falling object, heavy or light, picks up speed at the same rate. That is the fair version of the rule Aristotle got wrong. Gravity is also the reason a ball thrown upward can meet the same height twice, once going up and once coming down. When it does, both of those roots are positive, and this is the lesson where you learned to keep them.