Lesson 10.4: Solve Applications Modeled by Quadratic Equations

New Concept

This lesson empowers you to translate real-world scenarios—like problems involving area, geometry, or projectile motion—into quadratic equations. You will apply various algebraic methods to find practical solutions to these complex, real-life applications.

What’s next

Next, you'll tackle interactive examples that break down the problem-solving strategy, followed by a series of practice cards to build and test your new skills.

Property

Consecutive even or odd integers can be represented algebraically.
If we call the first integer $n$, the next consecutive even or odd integer is $n+2$, the next is $n+4$, and so on.
This allows us to model problems about their products with quadratic equations.

Consecutive even integers: $n, n+2, n+4, \dots$
Consecutive odd integers: $n, n+2, n+4, \dots$

Examples

• The product of two consecutive odd integers is 143. Let the integers be $n$ and $n+2$. The equation is $n(n+2) = 143$, which simplifies to $n^2 + 2n - 143 = 0$. Factoring gives $(n+13)(n-11)=0$, so $n=11$ or $n=-13$. The pairs are 11, 13 and -13, -11.
• The product of two consecutive even integers is 288. Let them be $n$ and $n+2$. The equation is $n(n+2) = 288$, or $n^2 + 2n - 288 = 0$. Factoring gives $(n+18)(n-16)=0$, so $n=16$ or $n=-18$. The pairs are 16, 18 and -18, -16.
• The product of two consecutive odd integers is 399. Let them be $n$ and $n+2$. The equation is $n(n+2) = 399$, or $n^2 + 2n - 399 = 0$. Using the quadratic formula, we find $n=19$ or $n=-21$. The pairs are 19, 21 and -21, -19.

Property

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

Examples

• A triangle's area is 35 square feet. Its base is 3 feet more than its height. Let height be $h$ and base be $h+3$. The equation is $35 = \frac{1}{2}(h+3)h$, or $h^2+3h-70=0$. Solving gives $h=7$. The height is 7 ft and the base is 10 ft.
• A triangular banner has an area of 80 square inches. The height is 4 inches less than twice the base. Let base be $b$. The equation is $80 = \frac{1}{2}b(2b-4)$, or $b^2-2b-80=0$. Solving gives $b=10$. The base is 10 in and the height is 16 in.
• The area of a triangular garden plot is 50 square meters. The base is four times the height. Let height be $h$. The equation is $50 = \frac{1}{2}(4h)h$, or $2h^2=50$. This gives $h^2=25$, so $h=5$. The height is 5 m and the base is 20 m.

Explanation

This formula connects a triangle's area, base, and height. When one dimension is described in terms of the other, substituting into the formula creates a quadratic equation you can solve to find the exact lengths of the base and height.

Property

In any right triangle, where $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.

Examples

• A right triangle has legs of length $x$ and $x+7$. The hypotenuse is 13. The Pythagorean Theorem gives $x^2 + (x+7)^2 = 13^2$. This simplifies to $x^2+7x-60=0$. Solving gives $x=5$. The legs are 5 and 12.
• The hypotenuse of a right triangle is 25 cm. One leg is 5 cm shorter than the other. Let the legs be $x$ and $x-5$. The equation is $x^2 + (x-5)^2 = 25^2$, which simplifies to $x^2-5x-300=0$. Solving gives $x=20$. The legs are 20 cm and 15 cm.
• A ladder leans against a wall. The ladder is 17 feet long. The base of the ladder is 8 feet from the wall. Let the height it reaches be $h$. The equation is $h^2 + 8^2 = 17^2$. This gives $h^2 = 289-64=225$, so $h=15$. The ladder reaches 15 feet up the wall.

Explanation

This famous theorem applies only to right triangles, creating a powerful link between the three sides. When side lengths involve variables, this formula allows you to set up and solve a quadratic equation to find the missing lengths.

Property

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

Examples

• A rectangular garden has an area of 176 square feet. Its length is 5 feet more than its width. Let width be $w$. The equation is $w(w+5)=176$, or $w^2+5w-176=0$. Solving gives $w=11$. The width is 11 ft and the length is 16 ft.
• The area of a rectangular patio is 250 square meters. The length is twice the width. Let width be $w$. The equation is $w(2w)=250$, or $2w^2=250$. This gives $w^2=125$, so $w = \sqrt{125} \approx 11.2$. The width is about 11.2 m and the length is about 22.4 m.
• A rectangular screen has an area of 90 square inches. Its width is 1 inch less than half its length. Let length be $L$. The equation is $L(0.5L-1)=90$, so $0.5L^2-L-90=0$. This gives $L^2-2L-180=0$. Solving gives $L \approx 14.4$. The length is about 14.4 in and the width is about 6.2 in.

Explanation

This formula helps find unknown dimensions of a rectangle. When length is expressed in terms of width, substituting into the area formula creates a quadratic equation. Solving it reveals the exact measurements for the length and width.

Property

The height in feet, $h$, of an object shot upwards into the air with initial velocity, $v_0$, after $t$ seconds is given by the formula:

Examples

• An object is launched upwards with an initial velocity of 96 ft/s. To find when it reaches 128 feet, we solve $128 = -16t^2 + 96t$. This simplifies to $t^2-6t+8=0$. Factoring gives $(t-2)(t-4)=0$. It reaches 128 ft at 2 seconds and 4 seconds.
• A ball is thrown up with an initial velocity of 80 ft/s. To find when it hits the ground, we set $h=0$ and solve $0 = -16t^2 + 80t$. Factoring gives $-16t(t-5)=0$. It hits the ground at $t=5$ seconds.
• A rocket has an initial velocity of 160 ft/s. To find when it is at a height of 384 feet, we solve $384 = -16t^2 + 160t$. This is $t^2 - 10t + 24 = 0$. Factoring gives $(t-4)(t-6)=0$. It is at 384 ft at 4 seconds and 6 seconds.

Explanation

This formula models an object's vertical path under gravity. By setting a target height for $h$ and using an initial velocity $v_0$, you create a quadratic equation to find the time $t$ it takes to reach that height.