a b c 0 then roots of ax 2 bx c 0 is

x(ax+b)+c=0
x(ax-a-c)+c=0
ax^2-ax-cx+c=0
ax(x-1)-c(x-1)=0
(ax-c)(x-1)=0
thus x=c/a and x=1 are the roots
10 years agoHelpfull: Yes(29) No(2)
above answer submitted by Jayalakshmi.S is right but too lengthy so
you can use heat and trail method...
by putting x=1 you can get in eqn ax^2+bx+c=0 directly a+b+c=0
10 years agoHelpfull: Yes(24) No(3)
ansr will be= 1 root=(a-b-c)/2a
and other root=(-a-b+c)/2a

if u hav remembrd d option plz verify
10 years agoHelpfull: Yes(6) No(1)
x=(-b+-sqrt(b^2-4ac))/2a
x=(-b+_sqrt((a-c)^2)/2a
x=-(a+b+c)/2a and x=(-b+a+c)/2a
x=0 and x=1
10 years agoHelpfull: Yes(6) No(16)
since a+b+c=0;b=-(c+a);
root of equation=(-b+/-sqrt(b^2-4*a*c))/2*a;
after putting the value of b,we get
two roots=1,c/a;
10 years agoHelpfull: Yes(6) No(1)
as the roots of quadratic equation is given by
[-b+(b^2-4ac)]/2 and [-b-(b^2-4ac)]/2
and since a+b+c=0
b=-a-c
putting this value of b in the above two eq of roots we get
x1=1/2 and x2=c/a
10 years agoHelpfull: Yes(5) No(3)
the roots are...
(-b (+ or -) under root of b^2-4ac) /2a
10 years agoHelpfull: Yes(1) No(1)
solution:
roots are 1 and -1-b/a
10 years agoHelpfull: Yes(0) No(1)
If you check the answer by putting the options.
If we put the X=1 ans we will get a+b+c=o...
8 years agoHelpfull: Yes(0) No(0)

a+b+c=0, then roots of ax^2+bx+c=0 is?