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

[utkarshakash]

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?

 

[HallsofIvy]

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.

 

[Chestermiller]

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.

 

[utkarshakash]

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!