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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
maganbhai P.
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#

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