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If the roots of a quardiatic equation x^3-px^2 +qx-r =0 are in Arithmetic Progression (AP) Then which is the correct relation -
A) 2p^3=9pq-27r
B) 2q^3=9pq-27r
C) p^3 = 9pq-27r
D) 2p^3=9pq+27r

1 Answer
Aug 3, 2018

(A) :2p^3=9pq-27r

Explanation:

We have cubic eqn.

x^3-px^2+qx-r=0...........to(1)

Let ,a,b,c be the three roots of eqn. (1)

So ,

color(red)(a+b+c=-(-p)/1=>a+b+c=pto(2)

color(blue)(ab+bc+ca=q/1=>ab+bc+ca=qto(3)

andcolor(brown)( a*b*c=-(-r)/1=>a*b*c=rto (4)

Given that , a,b,c " are in A.P."

=>b-a=c-b=>a+c=2b...to(5)

From (2) and (5)

a+c+b=p=>2b+b=p=>3b=p=>b=p/3to(6)

From (3) and (5)

ab+bc+ca=q=>b(a+c)+ca=q

=>b(2b)+ca=qto[becausea+c=2b]

=>2b^2+ca=q

=>ca=q-2b^2.......... to[where ,b=p/3]

=>ca=q-(2p^2)/9....to(7)

From (4)

a*b*c=r

=>b*ac=r

Using (6)and (7)

p/3(q-(2p^2)/9)=r

=>(pq)/3-(2p^3)/27=r

=>9pq-2p^3=27r

=>9pq-27r=2p^3

i.e. 2p^3=9pq-27r

Note :
If alpha, beta and gamma are roots of cubic eqn.

Ax^3+Bx^2+Cx+D=0 ,then

(i)color(red)(alpha+beta+gamma=-B/A

(ii)color(blue)(alpha*beta+beta*gamma+gamma*alpha=C/A

(iii)color(brown)(alpha*beta*gamma=-D/A