Question

Let $\alpha ,\beta$ be the roots of $x^2-x+p=0$ and $\gamma ,\delta$ be roots of $x^2-4x+q=0$. If $\alpha ,\beta ,\gamma ,\delta$ are in GP, then integral values of $p$ and $q$ respectively are:

• A. $-2,-32$
• B. $-2,3$
• C. $-6,3$
• D. $-6,-32$

Answer 1

The correct option is A

$-2,-32$

Explanation for the correct option:

Finding the integral values of $p$ and $q$

Step 1: Determining the roots

Given $\alpha ,\beta$ are the roots of $x^2-x+p=0$ then

As we know that for the quadratic equation $a{x}^2+bx+c=0$

Sum of roots, $=-\frac{b}{a}$

Product of roots$=\frac{c}{a}$

Now comparing the above two equation we have $a=1,b=-1\&c=p$

Therefore, the sum of roots is

$$\begin{gathered} \Rightarrow \alpha +\beta =\frac{-b}{a}=\frac{-\left(-1\right)}{1} \\ \Rightarrow \alpha +\beta =1\dots \left(i\right) \end{gathered}$$

And product of roots is

$$\begin{gathered} \Rightarrow \alpha \beta =\frac{c}{a}=\frac{p}{1} \\ \Rightarrow \alpha \beta =p \end{gathered}$$

And also given $\gamma ,\delta$ are the roots of $x^2-4x+q=0$ then

Similarly,

Sum of roots,$\gamma +\delta =\frac{-\left(-4\right)}{1}=4\dots \left(ii\right)$

Product of roots, $\gamma \delta =\frac{q}{1}=q$

Given $\alpha ,\beta ,\gamma ,\delta$ are in GP.

Therefore, considering

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

$$\begin{gathered} \alpha +\beta =a+ar=1\dots .\left(iii\right)\mathbf{}\left[\text{from}\left(i\right)\right] \\ \gamma +\delta =a{r}^2+a{r}^3=4\mathbf{}\left[\text{from}\left(ii\right)\right] \\ \Rightarrow r^2(a+ar)=4 \\ \Rightarrow r^2\left(1\right)=4\left[\text{from}\left(iii\right)\right] \\ \Rightarrow r=\pm 2 \end{gathered}$$

Substituting $r$ in $\left(iii\right)$

Case 1: When $r=-2$

$$\begin{gathered} \Rightarrow a(1-2)=1 \\ \Rightarrow a=-1 \end{gathered}$$

Case 2: When $r=2$,

$$\begin{gathered} \Rightarrow a(1+2)=1 \\ \Rightarrow a=\frac{1}{3} \end{gathered}$$

Using both values of $a$ and $r$.

Step 2: Finding values of $p$

$$\begin{gathered} p=\alpha \beta \\ =a\left(ar\right) \\ =a^{2}r \\ \text{When}r=-2\text{and}a=-1 \\ ={\left(-1\right)}^2(-2) \\ =-2 \\ \text{When}r=2\text{and}a=\frac{1}{3} \\ p=a\left(ar\right) \\ ={\left(\frac{1}{3}\right)}^2\left(2\right) \\ =\frac{2}{9} \end{gathered}$$

Step 3: Finding values of $q$

$$\begin{gathered} q=\gamma \delta \\ =a{r}^{2}a{r}^3 \\ =a^2{r}^5 \\ \text{When r}=-2\text{and}a=-1 \\ ={\left(-1\right)}^2{\left(-2\right)}^5 \\ =-32 \\ \text{When}r=2\text{and}a=\frac{1}{3} \\ q=a^2{r}^5 \\ ={\left(\frac{1}{3}\right)}^2{\left(2\right)}^5 \\ =\frac{32}{9} \end{gathered}$$

Thus, the value of $p,q$ are $-2,-32$ respectively or $\frac{-2}{9},\frac{-32}{9}$respectively.

And $-2,-32$ is given in the options.

Hence, option (A) is the correct answer.