Greeings,

The equation below is given and the problem is to find all values of m, for which the roots of the equation form an arithmetic progression.

Though, I am pretty sure Vieta's formulae are to be used, I am puzzled by the fact they are usable in quadratic equations only..
I've considered a substitution of the x^2=y like, but then I loose the link between the substituted equation's roots and the biquadratic equation's roots..

Thank you.

2. Originally Posted by Logic
Greeings,

The equation below is given and the problem is to find all values of m, for which the roots of the equation form an arithmetic progression.

Though, I am pretty sure Vieta's formulae are to be used, I am puzzled by the fact they are usable in quadratic equations only..
I've considered a substitution of the x^2=y like, but then I loose the link between the substituted equation's roots and the biquadratic equation's roots..

Thank you.

$\displaystyle x^4-(3m+5)x^2+(m+1)^2=0$

Assume this has four distinct real roots (as it must if they are to be in arithmetic progression since it can only have 0, 2 or 4 real roots and so you need 4 to have an arithmetic progression), and as this is a quadratic in $\displaystyle y=x^2$, they are of the form: $\displaystyle -b,\ -a,\ a,\ b;\ \ b > a > 0$.

So for these to be in arithmetic progression means that:

$\displaystyle a=\frac{1}{2}(b-a)$

where $\displaystyle a$ and $\displaystyle b$ are the distinct positive roots of $\displaystyle y^2-(3m+5)y+(m+1)^2=0$ and $\displaystyle a$ is the smaller of the two.