1)If the equation x2+bx+ac=0 and x2+cx+ab=0 have a common root then prove that the equation containing other roots will be given by x2+ax+bc=0.
2)If the equations ax2+bx+c=0 and x3+3x2+3x+2=0 have two common roots then show a=b=c.
23051>
let m and n be the roots of the 1st equation and
let m and p be the roots of the 2nd
mn = ac
mp=ab
→np=cb....(1)
m+n = -b
m+p = -c
→p-n= b-c...(2)
From (1) and (2) we get
p=b and n=c
m = -n-b = abp=abb=a
→a= -n-b = -b-c
Equation containing n and p-
x2+(-n-p)x+np
=x2+ax+bc
23052>
x3+3x2+3x+2=0
→(x+1)3=-1
→x = -2,-1-w,-1-w2(where w is the cube root of unity)
Its known that for real a,b,c complex roots occur in conjugate pairs.
Therefore roots of the equation ax2+bx+c =0 are -1-w and -1-w2
-ba=-1-w-w2-1= -1(Because 1+w+w2=0)
→a=b
ca=(1+w)(1+w2)= 1+w+w2+w3= w3 = 1
→a=c
Hence,
a=b=c
