If the roots of the equation, 8 x^3+6 p x^2+3 q x-27=0 are in a geometric progression, then q^2+9 p^2+6 p q+q / p=

Step-by-step solution

Let roots of the equation 8 x^3+6 p x^2+3 q x-27=0 are a r , a, ar. a r +a+a r= -6 p 8 = -3 p 4 a ( 1 r +1+r )= -3 p 4 (i) ( a r a )+(a a r)+ ( a r a r )= 3 q 8 a^2 ( 1 r +r+1 )= 3 q 8 (ii) and a r a a r= 27 8 a^3= 27 8 a= 3 2 (iii) From Eqs. (i) and (ii), we get a= 3 q / 8 -3 p / 4 =- q 2 p (iv) From Eqs. (iii)and (iv), we have -q 2 p = 3 2 -q=3 p Now, q^2+9 p^2+6 p q+ q p 9 p^2+9 p^2+6 p(-3 p)+ (-3 p) p 18 p^2-18 p^2-3=-3