be the roots of; 
and 
be the roots of 
. If 
are in G.P., then the integer values of p and q respectively are:
A) -2, -32
B) -2, 3
C) -6, 3
D) -6. -32
Correct Answer: A
Solution :
Consider, eqn \[{{\operatorname{x}}^{2}}- x +p =0\] \[\left. \begin{matrix} \operatorname{a}+\beta =1 \\ \alpha \beta =p \\ \end{matrix} \right\}\] Now consider equation \[{{\operatorname{x}}^{2}}- 4x+q =0\] \[\left. \begin{matrix} \gamma +\delta =4 \\ \gamma \delta =q \\ \end{matrix} \right\}\] Let r be the common ratio. Then \[\operatorname{a}\,+\alpha r=1\Rightarrow \alpha (1+r)=1\] ... (i) and \[{{\operatorname{ar}}^{2}}+a{{r}^{3}}=\,\,4\,\,or\,\,a{{r}^{2}}\,(1+r)=4\] ? (ii) From (i) and (ii), \[{{\operatorname{r}}^{2}}= 4 \Rightarrow \,r = \pm 2\] Now, \[\alpha .\alpha r=p\,\,and\,\,\alpha {{r}^{2}}. \alpha {{r}^{3}}=q\] Putting \[\operatorname{r} = - 2\] we get \[\operatorname{a}=-1,\,\,p=-2\,\,and\,\,q=-32.\] Again \[r=2\,gives\,\,\alpha =1/3,\,\,and\,\,p=\frac{2}{9}.\] But p is an integer, therefore, \[\operatorname{p}= - 2\] and \[\operatorname{q} = -32\].
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