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 $G.P$, then find the value of $A+B$.

Answer 1

$x^2-8x+A=0$

$\alpha +\beta =8$...(1)

$\alpha \beta =A$...(2)

$x^2-72x+B=0$

$\gamma +\delta =72$...(3)

$\gamma \delta =B$...(4)

$\beta =\alpha r,\gamma =\alpha r^2,\delta =\alpha r^3$

From (1) $\alpha +\alpha r=8\Rightarrow \alpha (1+r)=8\Rightarrow 1+r=\frac{8}{\alpha }$...(5)

From (2) $\alpha \beta =A\Rightarrow {\alpha }^{2}r=A$...(6)

From (3) $\gamma +\delta =72\Rightarrow \alpha r^2+\alpha r^3=72$

$\alpha r^2(1+r)=72$

$\alpha r^2\left(\frac{8}{\alpha }\right)=72$

$8r^2=72\Rightarrow r^2=9\Rightarrow r=3$

From (4)

$\gamma \delta =B$

$\left(\alpha r^2\right)\left(\alpha r^3\right)=B$

${\alpha }^2{r}^5=B$

$243{\alpha }^2=B$

$81\times 3{\alpha }^2=B$

$81r{\alpha }^2=B$

$81A=B$

$\therefore A+B$

$=A+81A$

$=82A$