Question

Let $\alpha$ and $\beta$ be the roots of $x^2-3x+p=0$ and $\frac{1}{\gamma }$ and $\frac{1}{\delta }$ be the roots of $x^2-6x+q=0$. If $\alpha ,\beta ,\gamma ,\delta$ form a geometric progression. Then 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

$9:7$

Explanation for the correct option:

Step 1: Use relation between roots and coefficients of the quadratic equation

The given quadratic equation is $x^2-3x+p=0$, $\alpha \&\beta$ are the roots.

Therefore,

Sum of roots, $\alpha +\beta =3$

Product of roots, $\alpha \beta =p$

Also the in given quadratic equation $x^2-6x+q=0$, $\gamma \&\delta$ are roots.

therefore,

Sum of roots $\gamma +\delta =6$

Product of roots $\gamma \delta =q$

Since $\alpha ,\beta ,\gamma ,\delta$ form a geometric progression

Step 2: Find the required ratio

Therefore, consider $\alpha =a,\beta =ar,\gamma =a{r}^2,\delta =a{r}^3$

$$\begin{gathered} a(1+r)=3....\left(i\right)\left[\because \alpha +\beta =3\right] \\ a{r}^2(1+r)=6..\dots \left(ii\right)\left[\because \gamma +\delta =6\right] \end{gathered}$$

Dividing $\left(ii\right)$ by $\left(i\right)$

$$\begin{gathered} \Rightarrow \frac{a{r}^2(1+r)}{a(1+r)}=\frac{6}{3} \\ \Rightarrow r^2=2 \\ \because \alpha .\beta =p \\ =a^{2}r \\ \because \gamma .\delta =q \\ =a^2{r}^5 \\ \therefore \frac{(2q+p)}{(2q-p)}=\frac{(2r^4+1)}{(2r^4-1)} \\ =\frac{(2\times 2^2+1)}{(2\times 2^2-1)} \\ =\frac{9}{7} \end{gathered}$$

Thus, $(2q+p):(2q-p)=9:7$

Hence, the correct option is (B)