What Is the Probability of No Real Roots in a Quadratic Equation?

  • Thread starter Thread starter yuqiao
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
The discussion centers on calculating the probability that the quadratic equation x^2 + ax + b = 0 has no real roots when a and b are randomly selected from the interval [0, 2]. The key condition for the equation to lack real roots is that the discriminant, given by a^2 - 4b, must be less than zero. Participants are encouraged to share their attempts and specific challenges to facilitate assistance. The conversation emphasizes collaborative problem-solving in understanding the mathematical concepts involved. Engaging with the problem can lead to a clearer grasp of quadratic equations and their properties.
yuqiao
Messages
7
Reaction score
0

Homework Statement


Given an interval [0,2], randomly select two numbers a, b in it. What is the probability of the situation where the equation x^2+ax+b=0 has no real root?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Hi yuqiao! :smile:

(try using the X2 tag just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
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...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top