Obtaing a desired quadratic equation

AI Thread Summary
To obtain an inverted quadratic equation with a maximum of y=100 and roots at x=0 and x=1/120, the equation should be in the form y = A(x - 0)(x - 1/120). The constant A must be determined to ensure the maximum value is 100. The vertex of the parabola, where the maximum occurs, is located halfway between the roots. Understanding the relationship between the roots and the vertex is crucial for solving similar problems in the future. This approach will lead to the desired quadratic equation.
tsumi
Messages
15
Reaction score
0

Homework Statement



Get an inverted quadratic equation (-x^2) with maximum y=100 and roots x=0 and x=1/120.

Homework Equations





The Attempt at a Solution



I tried changing the several terms of the quadratic equation, used the quadratic formula setting c=100, and x=0 with the negative root and x=1/120 with the positive root, among other less smart attempts. Could not get something reasonable.

Does anyone know how to do it? It does not seam that much difficult, maybe my academic formation is just weak.. =\
 
Physics news on Phys.org
Hi tsumi,

You need to find the function y(x)=ax^2+bx+c, which crosses the x-axis at points x1=0 and x2=1/120, and its maximum is 100=y(max). Or you can write out the function in the form y=C-a(x-b)^2. In this case C=100 as you said, and the maximum is at x=b. Where is the position of the peak of a parabola with respect to its zero points?

ehild
 
Last edited:
tsumi said:

Homework Statement



Get an inverted quadratic equation (-x^2) with maximum y=100 and roots x=0 and x=1/120.

Homework Equations


The Attempt at a Solution



I tried changing the several terms of the quadratic equation, used the quadratic formula setting c=100, and x=0 with the negative root and x=1/120 with the positive root, among other less smart attempts. Could not get something reasonable.

Does anyone know how to do it? It does not seam that much difficult, maybe my academic formation is just weak.. =\

What you seem to be missing (I was not able to understand what you did) is that with roots , α, β, ... the equation
(x - α)(x - β) = 0 is satisfied. And so on for as many roots as there are. This is coming at it from the solved side so to speak - you have probably had more emphasis from the unsolved side.

The conditions you are required to satisfy involve more than the roots, in fact you need an equation of form

A(x - α)(x - β) = 0,

where A is a constant. You are being asked then also to find the A that gives you the maximum stated. If you revise (or work it out) you will find that an extremum for a quadratic is localised halfway between any real roots; however do not just blindly apply that without understanding where that comes from, certainly in your book, or work out self, otherwise you will not have benefited and made self able to solve next time.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top