Solve this tough GEOMETRIC PROGERSSION QUESTION..???

Question

LET a,b,c,d are Real Numbers in GP and p= -1/3 q= 2/3 r= 5/3

then,

(1/p + 1/q + 1/r)x^2 + ( (b-c)^2 + (c-a)^2 + (d-b)^2 )x + (p + q + r) = 0

and,

20x^2 + [10(a-d)^2]x - 9 =0

Prove that the ROOTS OF THESE TWO EQUATIONS ARE RECIPROCAL OF EACH OTHER.

Answer 1

For ax^2+bx+c=0 say

products of roots = c/a

here take first equation and find c/a say roots are 'm' and 'n'

means find (p+q+r)/ [(1/p + 1/q +1/r]

p+q+r=-1/3+2/3+5/3= (-1+2+5)/3=6/3=2

1/p + 1/q + 1/r = -3 +3/2 + 3/5= (-30+15+6)/10= -9/10

now  products of the roots=mn=(2)/(-9/10)= (- )20/9

for second equation (c/a)

= -9/20= (1/mn)=(1/m)(1/n)

so the  roots are reciprocal to each other