Let $a,b,c,d$ be real numbers in G.P.If $u,v,w$ satisfy the system of equations
$u+2v+3w=6$,
$4u+5v+6w=12$,
$6u+9v=4$
then show that the roots of the equation $$\left(\frac{1}{u}+\frac{1}{v} + \frac{1}{w} \right)x^2 + \left[(b-c)^2+(c-a)^2+(d-b)^2\right]x+u+v+w=0$$
and $20x^2+10(a-d)^2x-9=0$ are reciprocals of each other.
I am stuck in how to implement the linear equations in determining about the roots of the quadratic.Any hints or suggestion will be helpful.
Source:JEE 1999
On
HINT:
As $\dfrac ba=\dfrac cb=\dfrac dc;$ $b^2=ca,c^2=bd,ad=bc$
$(b-c)^2+(c-a)^2+(d-b)^2=2b^2+2c^2+a^2+d^2-2bc-2ca-2bd$ $=2(b^2-ca)+2(c^2-bd)+a^2+d^2-2ad=(a-d)^2$
Now
either $(i)$ use Cramer's Rule, to find $u,v,w$
or $(ii)$ use Cross-Multiplication Method on the first two given equations to express $u,v$ in terms of $w$ and then put those values in the third given equation.
Subracting first $u,v,w$ equation from second we get $u+v+w=2$.
We can easily solve to get $u=-\frac{1}{3},v=\frac{2}{3},w=\frac{5}{3}$. So $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=-\frac{9}{10}$.
Now suppose the roots of $$\left(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\right)x^2+[(b-c)^2+(c-a)^2+(d-b)^2]x+u+v+w=0\ \ (*)$$ are $\alpha,\beta$. Then we have $$\frac{1}{\alpha\beta}=\frac{\frac{1}{u}+\frac{1}{v}+\frac{1}{w}}{u+v+w}=-\frac{9}{10}\frac{1}{2}=-\frac{9}{20}$$ Which is indeed the product of the roots of $$20x^2+10(a-d)^2x-9=0$$
We also have $$\alpha+\beta=-\frac{(b-c)^2+(c-a)^2+(d-b)^2}{\frac{1}{u}+\frac{1}{v}+\frac{1}{w}}=\frac{10}{9}\left((b-c)^2+(c-a)^2+(d-b)^2\right)$$ so $$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-\frac{1}{2}\left((b-c)^2+(c-a)^2+(d-b)^2\right)$$ Thus we are done provided we can show that the last expression is $$-\frac{1}{2}(a-d)^2$$ or that $$(b-c)^2+(c-a)^2+(d-b)^2-(a-d)^2=0\ \ (**)$$ In general that is not true, but we are also given that $a,b,c,d$ is a geometric progression. So we have $b^2=ac,c^2=bd,bc=ad$. But expanding $(**)$ we get $(2b^2-2ac)+(2c^2-2bd)+(2ad-2bc)$ which is 0.