Question

Let $\alpha$ and $\beta$ be the roots of $x^2-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^2-6x+q=0$. If $\alpha ,\beta ,\gamma ,\delta$ form the geometric progression. Than ratio $(2q+p):(2q-p)$ is :

• A. $33:31$
• B. $9:7$
• C. $3:1$
• D. $5:3$

Answer 1

The correct option is B $9:7$

Let $\alpha$ and $\beta$ be the roots of $x^2-3x+p=0$
Then, $\alpha +\beta =3$ and $\alpha \beta =p$
and $\gamma$ and $\delta$ be the roots of $x^2-6x+q=0$
$\gamma +\delta =6$ and $\gamma \delta =q$
Since $\alpha ,\beta ,\gamma ,\delta$ are in G.P.
Let $\alpha =a,\beta =ar,\gamma =a{r}^2,\delta =a{r}^3$
Then, $\alpha +\beta =3\Rightarrow a(1+r)=3\cdots \left(1\right)$
$\gamma +\delta =6\Rightarrow a{r}^2(1+r)=6\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$r^2=2$
Now,
$\frac{(2q+p)}{(2q-p)}$
$=\frac{(2\gamma \delta +\alpha \beta )}{(2\gamma \delta -\alpha \beta )}=\frac{(2a^2{r}^5+a^{2}r)}{(2a^2{r}^5-a^{2}r)}$
$=\frac{(2r^4+1)}{(2r^4-1)}=\frac{2\cdot 2^2+1}{2\cdot 2^2-1}$
$=\frac{9}{7}$