quadratics - The value of $P^2+Q^2$ given that the reciprocals of the roots of two polynomials follow an arithmetic progression.

If $\alpha,\gamma$ are non zero roots of the equation $Px^2-4x+1=0$ and $\beta,\delta$ are non zero roots of $Qx^2-6x+1=0$ (where $\frac{1}{\alpha}$, $\frac{1}{\beta}$, $\frac{1}{\gamma}$ and $\frac{1}{\delta}$ are in arithmetic progression), then find the value of $(P^2+Q^2)$

I tried solving it but seem to be stuck. Any short solutions to this problem since it was given to us on a test?

Answer

Here is a solution using Vieta's formulas. Let $\alpha,\gamma$ be the roots of the polynomial $Px^2-4x+1$ and $\beta,\delta$ be the roots of the polynomial $Qx^2-6x+1$. Since we have an arithmetic progression, letting $1/\alpha=y$, we have that for some $d$:
$$1/\beta=y+d,\quad 1/\gamma=y+2d,\quad 1/\delta=y+3d$$
By Vieta's formulas, we have that:

$$\alpha+\gamma=\frac{4}{P},\qquad \alpha\cdot \gamma=\frac{1}{P} \tag{1}$$
$$\beta+\delta=\frac{6}{Q},\qquad \beta\cdot \delta=\frac{1}{Q} \tag{2}$$
This implies that:
$$\alpha+\gamma=4\alpha\cdot \gamma,\qquad \beta+\delta=6\beta\cdot \delta$$
From this, we have that:
$$\frac{1}{y}+\frac{1}{y+2d}=\frac{4}{y(y+2d)},\qquad \frac{1}{y+d}+\frac{1}{y+3d}=\frac{6}{(y+d)(y+3d)}$$
Hence:
$$\frac{(y+2d)+y}{y(y+2d)}=\frac{4}{y(y+2d)},\quad \frac{(y+3d)+(y+d)}{(y+d)(y+3d)}=\frac{6}{(y+d)(y+3d)}$$

This results in the system:

$$\begin{cases} 2y+2d=4 \\ 2y+4d=6 \end{cases}$$
Resulting in $y=1$ and $d=1$. So, the roots are $\alpha=1$, $\beta=1/2$, $\gamma=1/3$ and $\delta=1/4$. So then by equations $(1)$ and $(2)$, it follows that $P=3$ and $Q=8$. Thus, the answer is:
$$P^2+Q^2=3^2+8^2=73$$