Question

If $\alpha$ and $\beta$ are the roots of the equation$x^2-px+r=0$ and$\alpha /2,2\beta$are the roots of the equation$x^2-qx+r=0$. Then, the value of $r$ is

• A. $\frac{2}{9}(p–q)(2q–p)$
• B. $\frac{2}{9}(q-p)(2p–q)$
• C. $\frac{2}{9}(q–2p)(2q–p)$
• D. $\frac{2}{9}(2p–q)(2q–p)$

Answer 1

The correct option is D

$\frac{2}{9}(2p–q)(2q–p)$

Explanation for the correct option:

Step1. Expressing the given Quadratic equations :

Given that $\alpha$ and $\beta$ are the roots of the equation$x^2-px+r=0$

$\Rightarrow (\alpha +\beta )=p........\left(i\right);\alpha \beta =r.......\left(ii\right)$

And also given $\alpha /2,2\beta$are the roots of the equation$x^2-qx+r=0$.

$$\begin{gathered} \Rightarrow \frac{\alpha }{2}+2\beta =q;\frac{\alpha }{2}\times 2\beta =r \\ \Rightarrow \alpha +4\beta =2q;\alpha \beta =r \end{gathered}$$

$$\begin{gathered} \Rightarrow p+3\beta =2q \\ \Rightarrow 3\beta =2q–p \\ \Rightarrow \beta =\frac{(2q–p)}{3} \end{gathered}$$

Putting the value of $\beta$in equation (i). we get

$$\begin{gathered} \alpha +\frac{(2q–p)}{3}=p \\ \Rightarrow \alpha =\frac{(4p–2q)}{3} \end{gathered}$$

Step 2. Find the value of $\mathit{r}$:

Now, putting the value of $\alpha ,\beta$in equation (ii). we get

$$\begin{gathered} \alpha \beta =\frac{(2q–p)}{3}\frac{(4p–2q)}{3}=r \\ r=\frac{2}{9}(2p–q)(2q–p) \end{gathered}$$

Hence, the correct option is D.