Question

Let $\alpha ,\beta$ be the roots of $x^2-x+p=0and\gamma ,\delta$ be the roots of $x^2-4x+q=0.$~If $\alpha ,\beta ,\gamma ,\delta ,$ are in G.P., then the 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

$\alpha ,\beta$ are the roots of $x^2-x+p=0$
$\therefore \alpha +\beta =1...\left(i\right)\alpha \beta =p...\left(ii\right)$
$\gamma \delta \text{are the roots of}{x}^2+4x+q=0\therefore \gamma +\delta =4...\left(iii\right)\gamma \delta =q...\left(iv\right)\alpha ,\beta ,\gamma ,\delta \text{are in G.P.}$
$\therefore$ Let $\alpha =a;\beta =ar,\gamma =a{r}^2,\delta =a{r}^3.$
Substituting these values in equations (i),(ii),(iii) and (iv),
we get
$a+ar=1...\left(v\right)a^{2}r=p...\left(vi\right)a{r}^2+a{r}^3=4...\left(vii\right)a^2{r}^5=q...\left(viii\right)$
Dividing (vii) by (v) we get
$\frac{a{r}^2(1+r)}{a(1+r)}=\frac{4}{1}\Rightarrow r^2=4\Rightarrow r=2,-2\left(v\right)\Rightarrow a=\frac{1}{1+r}=\frac{1}{1+2}or\frac{1}{1-2}=\frac{1}{3}or-1$
As p is an integer (given), r is also an integer (2 or -2).
$\therefore \left(vi\right)\Rightarrow a\ne \frac{1}{3}.$
Hence $a=-1andr=-2.\therefore p=(-1{)}^2\times (-2)=-2q=(-1{)}^2\times (-2{)}^5=-32$