📘Intermediate Algebra Unit 9 Review

Quadratic equations show up constantly in practical problems: calculating areas, finding distances, and modeling motion. This section focuses on translating word problems into quadratic equations, solving them, and making sure your answers actually make sense in context.

Applications of Quadratic Equations

Most real-world quadratic problems fall into a few categories. Recognizing which type you're dealing with is the first step toward setting up the right equation.

Area problems involve rectangular or circular regions where dimensions are related to each other. For example, a garden's length might be 3 feet more than its width, and you're given the total area. If the width is $x$ , then the length is $x + 3$ , and the area equation becomes $x(x + 3) = 180$ .

Distance problems often use the Pythagorean theorem. A ladder leaning against a wall, or the diagonal of a rectangle, creates a right triangle. The theorem $a^2 + b^2 = c^2$ produces a quadratic when one or more sides are expressed in terms of a variable.

Product problems arise when two related quantities multiply to give a known result. Revenue problems are a common example: if you sell $x$ items at a price of $(50 - 2x)$ dollars each, revenue is $x(50 - 2x)$ , which is quadratic.

Setting Up and Solving

Read the problem carefully and identify what you're solving for. Assign a variable to the unknown quantity.
Express other quantities in terms of that variable using the relationships described in the problem.
Write an equation that connects everything (area formula, Pythagorean theorem, product relationship, etc.).
Solve the quadratic using whichever method fits best:
- Factoring works well when the equation factors neatly into two binomials
- Quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ works for any quadratic equation
- Completing the square is useful when you need exact answers or when the leading coefficient is 1
Check your solutions against the problem context. Reject any answer that doesn't make physical sense (negative lengths, negative time, etc.).
State your final answer with appropriate units.

Applications of quadratic equations, Solve Applications of Quadratic Equations · Intermediate Algebra

Quadratics in Uniform Motion

Uniform motion problems at this level typically involve objects traveling at different rates, where the relationship between their distances or times leads to a quadratic equation.

The core relationship is $d = rt$ , where $d$ is distance, $r$ is rate (speed), and $t$ is time. A quadratic appears when two quantities in this formula depend on the same variable.

Example setup: Two cyclists leave from the same point traveling at right angles. One rides 4 mph faster than the other. After 3 hours, they are 30 miles apart. Since they travel at right angles, the Pythagorean theorem applies to their distances, producing a quadratic in terms of the slower cyclist's speed.

When acceleration is involved, the distance equation becomes $d = v_0t + \frac{1}{2}at^2$ , where $v_0$ is initial velocity and $a$ is acceleration. This is already quadratic in $t$ , so you can substitute known values and solve directly.

Steps for Uniform Motion Problems

Draw a diagram if the problem involves directions or right angles.
Set up a table with columns for rate, time, and distance for each object or leg of the trip.
Express rate and time in terms of your chosen variable.
Use the problem's constraint (total distance, Pythagorean theorem, same arrival time, etc.) to write a quadratic equation.
Solve and check that your answers are realistic (positive speeds, positive times).

Quadratics for Projectile Motion

Projectile motion problems use a quadratic equation to describe the height of an object over time. The standard model for vertical height is:

$h = -16t^2 + v_0t + h_0$

where $h$ is the height (in feet) at time $t$ seconds, $v_0$ is the initial vertical velocity (in feet per second), and $h_0$ is the initial height. The $-16$ comes from half the acceleration due to gravity in feet ( $\frac{1}{2} \times 32 \text{ ft/s}^2$ ). In metric units, this constant is $-4.9$ (from $\frac{1}{2} \times 9.8 \text{ m/s}^2$ ).

Common questions you'll see:

When does the object hit the ground? Set $h = 0$ and solve for $t$ .
When does the object reach a certain height? Set $h$ equal to that height and solve for $t$ . You'll often get two answers: one on the way up, one on the way down.
What is the maximum height? Find the vertex of the parabola. The time at maximum height is $t = \frac{-b}{2a}$ , then plug that back in to find the height.

Solving a Projectile Problem Step by Step

Identify $v_0$ (initial velocity) and $h_0$ (launch height) from the problem.
Write the height equation with those values substituted in.
Set the equation equal to the target height (or 0 for ground level).
Rearrange into standard form $at^2 + bt + c = 0$ and solve.
Discard any negative time values. If you get two positive values, interpret both in context.

Graphing and Analyzing Quadratic Functions

The graph of a quadratic function is a parabola, and its features connect directly to the application problems above.

The vertex represents the maximum or minimum value. For projectile motion, this is the peak height. For revenue problems, it's the maximum revenue.
The x-intercepts (where the parabola crosses the horizontal axis) represent the solutions to the equation when the output equals zero. For projectile motion, these are the times when the object is at ground level.
If the parabola doesn't cross the x-axis, the equation has no real solutions, which means the scenario described in the problem doesn't occur (e.g., the object never reaches a specified height).

To find the vertex, use $x = \frac{-b}{2a}$ for the horizontal coordinate, then substitute back into the function for the vertical coordinate. This is the same technique used to find maximum height or maximum revenue in application problems.

📘Intermediate Algebra Unit 9 Review