Question

if the roots of the equation x^3+3px^2+3px+r=0 be in H.P then show that 2q^3=r(3pq-r).​

Answer 1

Answer:

If the roots of the equation

f(x) = x^3 + 3px^2 + 3qx + r = 0 are in H.P. then the roots of the equation f(1/x) = 0 will be in A.P. But f(1/x) = 0 is (1/x^3)+3p(1/x^2)+ 3q(1/x) + r =0 or

r x^3 + 3qx ^2 + 3px +1 = 0. Suppose

(a-d), a & (a+d) be the roots(taken in A.P.) Then by theory of equations, we get,(I) (a-d) + a + (a+d) = -3q/r ==> a = -q/r , (ii) (a-d)a(a+d) =-1/r ==> (a^2 -d^2)

=(-1/r)(-r/q) = 1/q ==> d^2 = a^2 - (1/q) = (q^2/r^2 -1/q) and

(iii) a(a-d)+a(a+d)+(a^2- d^2) = 3p/r ==> a{a-d + a+d}=(3p/r)-(1/q) or

2a^2 = (3pq -r)/qr ==>

2(q/r)^2 = (3pq-r)/qr , which in turn gives 2q^3 = r(3pq-r).