IIT-JEE 1999Question 1
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 $\left(\dfrac1u+\dfrac1v+\dfrac1w\right)x^{2}+\big[(b-c)^{2}+(c-a)^{2}+(d-b)^{2}\big]x+u+v+w=0$ and $20x^{2}+10(a-d)^{2}x-9=0$ are reciprocals of each other.
Sequence and Series · Geometric Progression
Answerproved
Solution
Formulas used. (1) For $a,b,c,d$ in G.P.: $b^2=ac$, $c^2=bd$, $bc=ad$(2) Replacing $x$ by $\frac1x$ in a quadratic gives the equation whose roots are the reciprocalsStep 1 - Solve the system. (Second) $-2\times$(first): $2u+v=0$, so $v=-2u$. Then $6u-18u=4$: $u=-\tfrac13$, $v=\tfrac23$, and $3w=6-u-2v=5$: $w=\tfrac53$. $$\frac1u+\frac1v+\frac1w=-3+\frac32+\frac35=-\frac9{10},\qquad u+v+w=2 .$$ Step 2 - The G.P. bracket. Expand and use formula (1): $$(b-c)^2+(c-a)^2+(d-b)^2=a^2+d^2+2(b^2+c^2)-2(bc+ac+bd)=a^2+d^2-2ad=(a-d)^2 .$$ Step 3 - First equation. $$-\frac9{10}x^2+(a-d)^2x+2=0 .$$ Step 4 - Reciprocal roots. Put $x\to\frac1x$ and multiply by $x^2$: $$2x^2+(a-d)^2x-\frac9{10}=0\ \Longleftrightarrow\ 20x^2+10(a-d)^2x-9=0,$$ which is the second equation. So the roots are reciprocals of each other. Answer: proved