GP problem w.r.t roots

Question

The sum of the squares if three distinct real number which are in GP is $S^2$. If their sum is $\alpha$S. Then$\alpha^2$ lies between

A) (1/3,2) B) (1,2) C). (1/3,3). D) none of these

Answer 1

Direction of Approach:

Let the three terms in geometric progression with common ratio $r$ and starting term $a$ be $\frac{a}{r}, a, ar$.

Given: $$\frac{a}{r}+a + ar = \alpha S \tag{1}$$ $$\frac{a^2}{r^2}+a^2+a^2r^2=S^2\tag2$$ Keeping that $r + \frac{1}{r}=y$, we can rewrite the equations as: $$a(y+1) = \alpha S \tag{1’}$$ $$a^2[(r^2+\frac{1}{r^2})+1]=S^2 \implies a^2[(r+\frac{1}{r})^2-1]= S^2$$ $$\implies a^2(y^2-1)= S^2 \tag{2’}$$

Dividing $(1’)$ and $(2’)$ and simplifying, we get, $$ y = \frac{\alpha^2+1}{\alpha^2-1} \implies r + \frac{1}{r}=\frac{\alpha^2+1}{\alpha^2-1}$$ $$\implies r^2(\alpha^2-1)-(\alpha^2+1)r+(\alpha^2-1)=0$$

Now, since $r$ is real, Discriminant $\geq 0 \implies \, ?$

Answer 2

WLOG $a^2+(ar)^2+(ar^2)^2=S^2$

$a+ar+ar^2=\alpha^2S$

$$\dfrac{a^2(1+r^2+r^4)}{a^2(1+r+r^2)^2}=\dfrac{S^2}{\alpha^2S^2}$$

$$\iff\alpha^2=\dfrac{1+r+r^2}{1-r+r^2}$$ as $1+r^2=r^4=(1+r^2)^2-r^2=?$

Rearrange to form Quadratic Equation in $r(\ne1)$ which is real, so the discriminant must be $\ge0$