Applications of Quadratics

Intuition

You can now solve any quadratic: you factor it when it factors, and you reach for the quadratic formula when it does not. 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: the algebra usually hands you two roots, and the situation almost always allows only one of them. A length cannot be negative, a clock cannot read a negative time, 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, factoring first if it factors, otherwise the formula. Then look at each 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 has two roots, and the equation cannot tell on its own which one 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.

Why an area or product problem keeps exactly one root#

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.

Recall the product-of-roots fact you proved for a quadratic with leading coefficient $1$ : the two roots multiply to the constant term. Here the constant term is $-A$ , so 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. The positive root is the one the situation keeps, and it is unique.

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 exactly one survives.

Projectile motion

Throw a ball straight up, or launch a rocket, and its height above the ground changes with time in a way physics describes with a quadratic. Near the surface of the Earth, an object in flight 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 $v_0$ 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 gravity is a little weaker in number, and the same model reads $h = -4.9t^2 + v_0\,t + h_0$ with the height in metres. 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.

There is one more thing the height model can tell you. Ask for a height the object never reaches, and the quadratic $h = H$ comes out with a negative discriminant. Since $\sqrt{-1} = i$ , 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.

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.

Worked example 3 A rectangle with area $40$

A rectangle is $3$ centimetres longer than it is wide, and its area is $40$ square centimetres. Find its dimensions.

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

The rectangle of Worked Example 3. Both sides are written with the same unknown w, so the area w(w + 3) = 40 is a quadratic. Solving gives w = 5 (the width) and w = -8, which is rejected because a length cannot be negative.

A right triangle brings in the Pythagorean theorem, $\text{leg}^2 + \text{leg}^2 = \text{hypotenuse}^2$ , which you met in pre-algebra. When the two legs are written with a shared unknown, squaring them produces a quadratic.

Worked example 4 A right triangle by the Pythagorean theorem

The longer leg of a right triangle is $7$ centimetres longer than the shorter leg, and the hypotenuse is $13$ centimetres. Find the two legs.

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

The right triangle of Worked Example 4. The legs share the unknown x, so the Pythagorean relation x squared plus (x + 7) squared equals 13 squared is a quadratic. Its roots are x = 5 (the shorter leg) and x = -12, which is rejected as a negative length.

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 $5$ centimetres” is the answer.

Key takeaways

Model an applied problem in four steps: name the unknown, then translate the words into a quadratic in standard form. Solve it (factor when it factors, otherwise use the quadratic formula), then interpret the roots and answer in a sentence with units.
The interpretation step is the point of the lesson: a quadratic gives two roots, and you keep only the ones the situation allows. Discard a negative length, a negative time, or a count that is not a positive whole number, and say why.
When the standard form has a positive leading coefficient and a negative constant term (area and product problems), the two roots have opposite signs. Exactly one positive root survives.
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 metres). Set $h = 0$ for “hits the ground” and $h = H$ for “reaches height $H$ ”; an object can pass a height twice, so both positive times may count.
Geometry gives quadratics through area, either a rectangle whose sides share one unknown or a border of width $x$ that adds $2x$ to each dimension. Geometry also gives quadratics through the Pythagorean theorem when the legs share an unknown.
A negative discriminant in a height problem is not an error: it reports that the object never reaches that height at all.

Applications of Quadratics

Learning goals

From words to a quadratic

Projectile motion

Geometry: area and right triangles

Number and money problems

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)