If roots of equation x^2-bx+c=0 be two consecutive integers, then b^2-4c equals?

• let the roots of the given equation be x and (x+1)=>two consecutive integers
  then sum of roots= (x+(x+1))=-b/a,(a=1 in the given equation)
  so (2x+1)=-b
  (-2x-1)=b => (1)
  product of roots= (x(x+1)=c/a,(a=1 in the given equation)
  so (x^2+x)=c => (2)
  therefore substituting the values of b and c in b^2-4c we get
  (-2x-1)^2-4(x^2+x)= 4x^2+4x+1-4x^2-4x = 1
  so b^-4c= 1
• 10 years agoHelpfull: Yes(3) No(0)
• if b=3 and c=2 then x^2-3x+2=0
  x^2-2x-x+2=0
  x(x-2)-1(x-2)=0
  x=1,2
  b^2-4c= 3^2-(4*2)=9-8=1
• 10 years agoHelpfull: Yes(0) No(0)
• x^2-bx+c===>two integer
  b^2-4c=====>two integers
• 10 years agoHelpfull: Yes(0) No(0)