Q. ax^2+bx+c = 0 has two roots x1 and x2. If mode x1 = mode x2, then
Option
(a) a>c
(b) b>c
(c) c>a
(d) none

• mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.
  
  in this case b=0
  eqn is ax^2+c = 0 => x=sqrt(-c/a)
  for x to be real (-c/a) should +ve => c & a are of opposite sign
  so c>a or a>c
  
  for ex x^2-4=0 => x1=-2,x2=2 & modx1=modx2=2 [a>c]
  or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 [c>a]
  
  (d) none
• 11 years agoHelpfull: Yes(29) No(3)
• c. c>a is the ans
• 11 years agoHelpfull: Yes(1) No(4)
• since x1*x2=c/a
  x1*x2= +VE ( x1=|x2| )
  c/a= +ve Const.
  so c=a(Const)
  hence c>a .......... similarly
  
  x1+x2= const= b/a
  b= a*const
  so b>a
  but we have c>a in option ...... it is the ans..
• 11 years agoHelpfull: Yes(1) No(3)
• i'm not sure but mod x1=mod x2 when b=0;then ax^2+c=0,x=underroot(-c/a),for this (-c/a) should be positive ,for this either c or a can be negative not both(for real roots)...so ans. is none
• 11 years agoHelpfull: Yes(1) No(2)
• |x1| = |x2|, means either both are same or both have opposite sign. Lets consder the first case. (x-2)^2 = x^2 - 2x + 4, a=1,b=-2,c=4. so if both root are same then. a is always 1,b is negative and c is always positive. so c>a.
• 11 years agoHelpfull: Yes(1) No(4)
• is it a ?? a>c ?
• 11 years agoHelpfull: Yes(0) No(5)