Question

Let ${\alpha }_1,{\beta }_1$ are the roots of $x^2-6x+p=0$ and ${\alpha }_2,{\beta }_2$ are the roots of $x^2-54x+q=0.$ If ${\alpha }_1,{\beta }_1,{\alpha }_2,{\beta }_2$ form an increasing G.P., then find the of the value of $(q-p).$

Answer 1

Let ${\alpha }_1=A,{\beta }_1=AR,{\alpha }_2=A{R}^2,{\beta }_2=A{R}^3$

we have ${\alpha }_1+{\beta }_1=6\Rightarrow A(1+R)=6$ ........ (1)

${\alpha }_1{\beta }_1=p\Rightarrow A^{2}R=p$

Also ${\alpha }_2+{\beta }_2=54\Rightarrow A{R}^2(1+R)=54$ ..........(2)

${\alpha }_2{\beta }_2=q\Rightarrow A^2{R}^5=q$

Now, on dividing Eq. (2) by Eq. (1), we get

$\frac{A{R}^2(1+R)}{A(1+R)}=\frac{54}{6}=9\Rightarrow R^2=9$

$\therefore R=3$ (As it is an increasing G.P.)

$\therefore$ On putting $R=3$ in Eq. (1), we get

$A=\frac{6}{4}=\frac{3}{2}$

$\therefore p=A^{2}R=\frac{9}{4}\times 3=\frac{27}{4}$ and $q=A^2{R}^5=\frac{9}{4}\times 243=\frac{2187}{4}$

Hence, $q-p=\frac{2187-27}{4}=\frac{2160}{4}=540$