Question

Let $\alpha_1, \beta_1$ are the roots of $x^2-6 x+p=0$ and $\alpha_2, \beta_2$ are the roots of $x^2-54 x+q=0$. If $\alpha_1, \beta_1, \alpha_2, \beta_2$ form an increasing G.P., then find the value of $(q-p)$.

Answer 1

$\text { Let } \alpha_1= A , \beta_1= AR , \alpha_2= AR ^2, \beta_2= AR ^3 $
$\text { we have } \alpha_1+\beta_1=6 \Rightarrow A (1+ R )=6$....(1)
$\alpha_1 \beta_1=p \Rightarrow A^2 R=p$....(2)
Also $\alpha_2+\beta_2=54 \Rightarrow \operatorname{AR}^2(1+ R )=54 $....(3)
$\alpha_2 \beta_2= q \,\,\,\, A ^2 R ^5= q$....(4)
Now, on dividing equation (3) by equation (1), we get
$\frac{ AR ^2(1+ R )}{ A (1+ R )}=\frac{54}{6}=9 $
$\Rightarrow R ^2=9 $
$\therefore R =3 \text { (As it is an increasing G.P.) } $
$\therefore \text { On putting } R =3 \text { in equation }(1) $
$A =\frac{6}{4}=\frac{3}{2}$
$\therefore p = A ^2 R =\frac{9}{4} \times 3=\frac{27}{4} $
$\text { and } q = A ^2 R ^5=\frac{9}{4} \times 243=\frac{2187}{4}$
$\text { Hence } q - p =\frac{2187-27}{4}=\frac{2160}{4}=540$