Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking for reasonably real, non-abstract applications modeled by quadratic equations where both solutions make sense. I'd like them to be accessible to high school algebra students.

One I come up with is firing a bullet through a sphere -- the two solutions correspond to the entry and exit points of the bullet, that is the intersections of the bullet's line and the sphere's surface. To keep it simple, we could do this instead with a line and circle in the xy plane, but the idea is the same.

Applications outside of physics would be particularly welcome.

share|improve this question
    
I suppose "where both solutions make sense" is in contrast to something like a right triangle where one leg is 7 less than the other and c = 13... my difficulty with this would be convincing students that some phenomenon (besides falling objects) exhibits quadratic behavior. –  The Chaz 2.0 Mar 14 '11 at 17:22

4 Answers 4

up vote 8 down vote accepted

Throw a ball upwards, you wanna know how long until it reaches some height?
Solve a quadratic, the two solutions correspond to when its on its way up and when its coming down again.

share|improve this answer

The simplest question I can think of is to characterize rectangles with side lengths $x, y$ and fixed perimeter $2x + 2y = p$ and area $xy = A$. The two roots of the corresponding quadratic equation (say in $x$), when they exist, correspond to the "tall" solution and the "wide" solution. But note that the statement of the problem is symmetric in $x$ and $y$, so if you know that a solution exists with $x \neq y$ then from one solution you automatically deduce the other. I think this is a nice example for students to see.

share|improve this answer
2  
How about where the perimeter of three sides is fixed (classic question: Rectangular goat pen which abuts the side of the barn and using a fixed amount of fence for the remaining three sides)? That breaks the symmetry. –  Ben Voigt Mar 14 '11 at 23:23

Knowing two sides $a$ and $c$ of a triangle and an angle $\alpha$ not between them—i.e. the SSA condition—does not determine the length of the third side $b$. See Figure 3 on this page. (Unfortunately, the convention at the length of the side opposite the vertex $A$ is denoted $a$ is going to make the notation a little confusing below.)

Set the origin at the vertex $A$ with the $x$-axis along $AC$. Then the vertex $B$ is at $(c \cos \alpha, c \sin \alpha)$ and $C$ is at $(b,0)$, and we have $$(c \cos \alpha - b)^2 + (c \sin \alpha - 0)^2 = a^{2},$$ which is a quadratic in $b$ both of whose solutions evidently make sense.

share|improve this answer
    
Very neat -- knew that but forgot. –  David Lewis Mar 14 '11 at 20:08

An application from number theory is elementary recurrence relations, with the Fibonacci numbers providing the easy example: the equation $F_{n+2} = F_{n+1}+F_n$ ties neatly into the quadratic equation $x^2-x-1=0$, and the Fibonacci numbers themselves turn out to be a linear combination of powers of the two solutions of this equation, $\phi$ and $-\phi^{-1}$. (The fact that the second solution is 'small' means that its contribution tends to $0$, but this isn't always the case; for instance, the equation $a_{n+2} = 5a_{n+1}-6a_{n}$ has as its 'core' solutions $2^n$ and $3^n$ and the general solution is a linear combination of these).

share|improve this answer
1  
This seems a little abstract to me. (I do love this application but I am not optimistic about explaining it to a typical class of high-schoolers.) –  Qiaochu Yuan Mar 14 '11 at 18:40
    
Yes, but I'm still liking it because, if you pose it as the golden ratio: a/b = b/(b+a), then the two solutions to x^2 + x - 1 = 0 are the golden ratio and its negative inverse. I think high schoolers could handle that. –  David Lewis Mar 14 '11 at 20:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.