Lesson 9.5: Solve Applications of Quadratic Equations

New Concept

This lesson empowers you to translate real-world scenarios—from geometry to projectile motion—into quadratic equations. You'll apply various solving methods to find practical solutions for these complex problems, bringing algebra to life.

What’s next

Next, you'll tackle these applications through worked examples and interactive practice cards, starting with problems involving consecutive integers and geometric shapes.

Property

For applications involving consecutive integers modeled by quadratic equations, the variables are defined as follows:

Consecutive even integers
$n$ = 1st even integer
$n+2$ = 2nd consecutive even integer
$n+4$ = 3rd consecutive even integer

Consecutive odd integers
$n$ = 1st odd integer
$n+2$ = 2nd consecutive odd integer
$n+4$ = 3rd consecutive odd integer

Property

For a triangle with base, $b$, and height, $h$, the area, $A$, is given by the formula:

To solve applications, let one dimension be a variable (e.g., $h$) and express the other dimension in terms of that variable (e.g., $b = 2h+4$). Substitute these into the area formula and solve the resulting quadratic equation.

Examples

• The area of a triangle is 40 square feet. Its base is 2 feet more than its height. Find the base and height. Let the height be $h$. The base is $h+2$. The equation is $40 = \frac{1}{2}(h+2)h$, which simplifies to $h^2 + 2h - 80 = 0$. Factoring gives $(h+10)(h-8)=0$. The height is 8 feet and the base is 10 feet.

• A triangle has an area of 78 square inches, and its base is 1 inch more than its height. Find its dimensions. Let the height be $h$. The equation is $78 = \frac{1}{2}(h+1)h$, or $h^2 + h - 156 = 0$. Factoring gives $(h+13)(h-12)=0$. The height is 12 inches and the base is 13 inches.

Property

For a rectangle with length, $L$, and width, $W$, the area, $A$, is given by the formula:

In application problems, one dimension is often expressed in terms of the other (e.g., $L = 3W-1$). Substituting this into the area formula creates a quadratic equation. If the equation cannot be factored, use the Quadratic Formula to find the solution and round as needed.

Examples

• A rectangular garden has an area of 100 square feet. Its length is 15 feet more than its width. Find the dimensions. Let the width be $W$. The equation is $100 = W(W+15)$, or $W^2 + 15W - 100 = 0$. Factoring gives $(W+20)(W-5)=0$. The width is 5 feet and the length is 20 feet.

• The length of a rectangular patio is 1 foot less than twice its width. The area is 105 square feet. Find the length and width. Let the width be $W$. The equation is $105 = W(2W-1)$, or $2W^2 - W - 105 = 0$. Factoring gives $(2W+15)(W-7)=0$. The width is 7 feet and the length is 13 feet.

Property

In any right triangle, where $a$ and $b$ are the lengths of the legs, and $c$ is the length of the hypotenuse, the relationship between the sides is given by the Pythagorean Theorem:

When side lengths are unknown but related to each other, substituting them into the formula creates a quadratic equation.

Examples

• One leg of a right triangle is 2 inches longer than the other leg, and the hypotenuse is 10 inches. Find the lengths of the legs. Let the legs be $x$ and $x+2$. The equation is $x^2 + (x+2)^2 = 10^2$, which simplifies to $x^2 + 2x - 48 = 0$. Factoring gives $(x+8)(x-6)=0$. The legs are 6 inches and 8 inches.

• The hypotenuse of a right triangle is 13 feet. The length of one leg is 7 feet more than the other leg. Find the side lengths. Let the legs be $x$ and $x+7$. The equation is $x^2 + (x+7)^2 = 13^2$, which becomes $x^2 + 7x - 60 = 0$. Factoring gives $(x+12)(x-5)=0$. The legs are 5 feet and 12 feet.

Property

Uniform motion problems involving a round trip against and with a current (like wind or water) can be modeled using the formula $t = \dfrac{D}{r}$. The total time is the sum of the time for each direction:

Multiplying by the LCD, $(r_{vehicle} - r_{current})(r_{vehicle} + r_{current})$, converts this rational equation into a quadratic equation.

Examples

• A plane flies a 1,200-mile trip each way. The round trip takes 10 hours with a plane speed of 250 mph. Find the wind speed. The equation is $\frac{1200}{250-r} + \frac{1200}{250+r} = 10$. This simplifies to $r^2 = 2500$, so the wind speed is 50 mph.

• A boat travels 15 miles upstream and back in 4 hours. The boat's speed in still water is 8 mph. Find the current's speed. We solve $\frac{15}{8-r} + \frac{15}{8+r} = 4$. This simplifies to $4r^2 = 4$, so the current's speed is 2 mph.

Property

For work applications where two individuals or machines work together, their combined rate is the sum of their individual rates. If they take $t_1$ and $t_2$ hours to do a job alone, the equation is:

This is a rational equation. To solve it, multiply by the LCD to eliminate the denominators, which results in a quadratic equation.

Examples

• Two painters can paint a room together in 6 hours. The experienced painter takes 9 hours less than the apprentice. How long does it take each painter alone? Let $t$ be the apprentice's time. $\frac{1}{t} + \frac{1}{t-9} = \frac{1}{6}$. This simplifies to $t^2 - 21t + 54 = 0$, so $(t-18)(t-3)=0$. The apprentice takes 18 hours, and the painter takes 9 hours.

• Two gardeners can landscape a yard in 4 hours together. The senior gardener is 6 hours faster than the junior gardener. Find the time for each. Let $t$ be the junior's time. $\frac{1}{t} + \frac{1}{t-6} = \frac{1}{4}$. This gives $t^2 - 14t + 24 = 0$, so $(t-12)(t-2)=0$. The junior gardener takes 12 hours, and the senior takes 6 hours.