Using properties of determinants, show that pα^2 + 2qα + r = 0, given that p, q and r are not in GP |(1,(q/p),α+(q/p)), (1, (r/q),α+(r/q))

Question

Using properties of determinants, show that pα² + 2qα + r = 0, given that p, q and r are not in GP 

\(\begin{bmatrix} 1& \frac{q}{p} & α+ \frac{q}{p} \\[0.3em] 1 &\frac{r}{q} & α +\frac{r}{q} \\[0.3em] p α+q & qα+ r & 0 \end{bmatrix} = 0\)

Answer 1

Given determinant

Given that p, q and r are not in GP

i.e., pr ≠ q²

∴ α²p + 2qα + r = 0 (proved).