Question

If $\alpha ,\beta$ are the roots of $x^2-8x+A=0$ and $\gamma ,\delta$ are the roots of $x^2-72x+B=0,$ if $\alpha <\beta <\gamma <\delta$ are in GP, then find the value of $A+B.$

Answer 1

Given eq. is $x^2-8x+A=0$...(1)

Since $\alpha ,\beta$ are roots of eq. (1)

sum of roots $\alpha +\beta =\frac{8}{1}=8$

Product of roots $\alpha \cdot \beta =\frac{A}{1}=A$

and $x^2-72x+B=0$...(2)

Since $\gamma ,\delta$ are roots of eq. (2)

$\gamma +\delta =72,\gamma .\delta =B$

Since $\alpha ,\beta ,\gamma ,\delta$ are in G.P.

Let common ratio be $\pi$

$\therefore \beta =d\cdot \pi ,\gamma =d\cdot {\pi }^2,\delta =d\cdot {\pi }^3$

$\alpha +\beta =8\Rightarrow \alpha +d\cdot \pi =8\Rightarrow \alpha (1+\pi )=8\Rightarrow 1+\pi =\frac{8}{\pi }$

$\alpha \cdot \beta =A\Rightarrow \alpha \cdot \alpha \pi =A\Rightarrow {\alpha }^2\cdot \pi =A$

$\gamma +\delta =72\Rightarrow \alpha \cdot {\pi }^2+alpha\cdot {\pi }^3=72\Rightarrow alpha\cdot {\pi }^2(1+\pi )=72$

$\Rightarrow alpha\cdot {\pi }^2\frac{8}{2}=72\Rightarrow \pi =3$

$\gamma \cdot \delta =B\Rightarrow \alpha \cdot {\pi }^2\cdot \alpha \cdot {\pi }^3=B$

$={\alpha }^2\cdot \pi \cdot {\pi }^4=B$

$=A\cdot 81=B$

$\therefore A:B=81:1$

$\therefore A+B=82$