Question

Let $x_1$ and $x_2$ be the roots of the equation $x^2-3x+A=0$ and let $x_3$ and $x_4$ be the roots of the equation $x^2-12x+B=0$. it is known that the numbers $x_1,x_2,x_3,x_4$ (in that order) form an increasing G.P. Find $\frac{B}{2A}$.

Answer 1

From the given conditions, we have $x_1+x_2=3$ and $x_3+x_4=12$
Also, $x_1,x_2,x_3,x_4$ form a G.P. and so we can write $x_1,x_2,x_3,x_4$ as $\frac{a}{r^3},\frac{a}{r},ar,a{r}^3.$
Now, $\frac{a}{r^3}+\frac{a}{r}=\frac{a}{r^3}(1+r^2)=3$ and $ar(1+r^2)=12$
Dividing the two, we have $r^4=4$ implying $r=\sqrt{2}$ since its an increasing G.P.
Thus, $a=2\sqrt{2}$
So, $A=x_1{x}_2=2$ and $B=x_3{x}_4=32$

So, $\frac{B}{2A}=8$.