Harmonic Mean of Roots: Solving a Quadratic Equation with Complex Terms

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Homework Help Overview

The discussion revolves around finding the harmonic mean of the roots of a quadratic equation with complex coefficients: (5+\sqrt{2})x^2-(4+\sqrt{5})x+8+2\sqrt{5}=0. Participants are exploring methods to approach the problem without directly calculating the roots.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest that finding the roots is not necessary for solving the problem and propose alternative methods, such as simplifying the equation or using properties of the roots. Others introduce the idea of substituting variables to relate the roots of the original equation to a new equation.

Discussion Status

The discussion is active with multiple participants offering different perspectives on how to approach the problem. There is a focus on simplifying the equation and utilizing relationships between the roots rather than calculating them directly. No consensus has been reached, but several viable approaches have been suggested.

Contextual Notes

Participants are navigating the complexities of the quadratic equation's coefficients and are considering the implications of their simplifications on the harmonic mean calculation. There is an emphasis on understanding the relationships between the roots rather than resolving the equation fully.

utkarshakash
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Homework Statement


The harmonic mean of the roots of the equation [itex](5+\sqrt{2})x^2-(4+\sqrt{5})x+8+2\sqrt{5}=0[/itex]

Homework Equations



The Attempt at a Solution



I know this question is easy but the main problem arises in finding the roots of the above equation. When I use the quadratic formula I get some complicated terms which is not easy to solve. What should I do?
 
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No, "finding the roots of the equation" is not the hard part because you don't need to find the roots! The first thing I would do is divide the entire equation by [itex]5+\sqrt{2}[/itex] to make the leading coefficient 1. Such a quadratic equation can be written as [itex](x- a)(x- b)= x^2- (a+b)x+ ab= 0[/itex] where a and b are the roots. You can read both a+ b and ab directly from the equation and use them to find the harmonic mean.
 
Substitute x = 1/y. Then the roots of the quadratic equation for y are the reciprocals of the roots of the equation for x. In the quadratic equation for y, -b/a is the sum of the roots for y, and is also equal to the sum of the reciprocals of the roots for x.
 
HallsofIvy said:
No, "finding the roots of the equation" is not the hard part because you don't need to find the roots! The first thing I would do is divide the entire equation by [itex]5+\sqrt{2}[/itex] to make the leading coefficient 1. Such a quadratic equation can be written as [itex](x- a)(x- b)= x^2- (a+b)x+ ab= 0[/itex] where a and b are the roots. You can read both a+ b and ab directly from the equation and use them to find the harmonic mean.

Thanks!
 

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