Applications of Quadratic Equations

Main idea

Area problems, falling objects and revenue models all lead to quadratics. A quadratic usually hands you two answers, so the last step is always asking which ones make sense in the real situation.

Height of a thrown object

h(t) = {\htmlData{sym=0}{h_0}} + {\htmlData{sym=1}{v_0}}t - \tfrac12{\htmlData{sym=2}{g}}t^2

What the symbols mean

$h_0$
Starting height
Where it's released.
$v_0$
Initial upward speed
How fast it's thrown up.
$g$
Gravity
9.8 m/s² (or 32 ft/s²).

Explanation

A quadratic application is simply a real situation whose equation has a squared term, and the most famous one is the formula up top for something thrown into the air. It has three pieces: $h_0$ (where it starts), $v_0t$ (how far its upward speed would carry it if nothing pulled it down) and $\frac12gt^2$ (how much gravity has pulled it back by time $t$ ). Gravity's effect grows with the square of time, which is exactly why a quadratic shows up. Area and revenue problems get their squares the same way, from multiplying two things that both depend on $x$ .

Why two answers? A parabola can cross a horizontal line twice, so a quadratic usually gives two solutions. The context decides which to keep. A negative time or length gets thrown out. But sometimes both answers are real: a ball passes any given height once on the way up and once on the way down, and both times are correct.

The turning point answers "best" questions. A parabola is symmetric, so its highest or lowest point sits exactly halfway between its two roots. Average the two roots from the quadratic formula and the $\pm$ parts cancel, leaving $x = -\frac{b}{2a}$ . So when a problem asks for the maximum height, the largest area or the best price, that's where to look. You'll meet this point again as the vertex in the Common Graphs unit.

See it

Here raising the price loses customers: at price p, 100 − 2p people buy, so revenue is R = p(100 − 2p), a downward parabola. Notice the target of 1200 is hit at two prices, 20 and 30, the two roots of a quadratic. Halfway between, at p = 25, revenue peaks at 1250.

Worked examples

Example 1

A ball is thrown up from 2 m at 15 m/s. When does it hit the ground? ( $g = 9.8$ )

1
Here $h_0 = 2$ , $v_0 = 15$ and $\frac12g = 4.9$ . It hits the ground when the height is zero: $2 + 15t - 4.9t^2 = 0$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A ball is thrown up from 2 m at 15 m/s ( $g = 9.8$ ). When does it land?

Problem 2+5 XP

A rectangle has area 60 and its length is 4 more than its width. Find the width.

Problem 3+5 XP

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

Problem 4+5 XP

For the ball in the first question, when is it highest?

Problem 5+5 XP

Revenue is $R = p(100 - 2p)$ . For which price is $R = 1200$ ? Give both.

Problem 6+5 XP

A farmer has 100 m of fence to make a rectangular pen along a straight river, with no fence needed on the river side. What's the largest area she can enclose, in m²?

Problem 7+5 XP

A ball's height is $h = 2 + 15t - 4.9t^2$ meters. What is its maximum height?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A 20 cm by 30 cm photo gets a frame of uniform width, and the framed picture has area 1000 cm². How wide is the frame, in cm?

Where it's used

Ballistics, braking distances (which grow with the square of speed) and pricing decisions all lead to quadratic equations.

Projectile MotionAlgebra · Physics

Completing the Square and the Quadratic Formula Equations Reducible to Quadratic Form