1. Find the roots a,b,c of x³-11x²+36x-36=0 if 2/b = 1/a + 1/c.

2. IF roots of x³+px²+qx+r=0 are in G.P. find the relations between p,q,r

you are given x^3 - 11 x^2 +36 x -36=0 Let us write (x+a) (x+ b) (x+c) Multiply them to get x^3 +(a+b+c) x^2 +( ab + ac+ bc) x - abc=0 comparing we get a+b+c = -11 (1) ab+ac+bc= 36 (2) abc = -36 (3) Divide (2) by (3). You get 1/c + 1/b + 1/a= -1 But we are given 1/a +1/c = 2/b When solved the above two equations, we get b=-3 Put this value in (1) and (2) You will get a+c=-8 ac= 12 Multiply first equation by c and subtract from 2. You will get c^2+8c+12= 0 (c+6)(c+2)=0 This give c= -6 or c=-2 Use c=-6 in (1). You will get a=-2