how many values of c in the equation x^2-5x+c result in rational roots which are intergers

• There are many values of C which will result in rational roots which are integers ( positive and negative).
  
  like c=6,-6,-14,-24,-50
• 12 years agoHelpfull: Yes(36) No(11)
• Infinite..
  c=4 => x=1,4
  c=-6 => x=-1,6
  c=-14 => x=-2,7
  c=-24 => x=-3,8
  c=-36 => x=9,-4
  c=-50 => x=-5,10
  c=-55 => x=-6,11 and so on...
• 12 years agoHelpfull: Yes(36) No(8)
• For roots of the equation X^2-5X+C to be rational integers we need to calculate its DISCRIMINANT (Dis) which should be greater than or equal to 0.
  i.e.
  
  Dis=b^2-4ac
  Dis>=0
  
  here b=-5 , a=1 , c=C
  
  hence Dis=(-5)^2 -4*1*C = 25-4C
  therefore we can write the inequality as
  
  25-4C>=0
  
  or
  25>=4C
  
  there are infinite values of C ranging from -infinity to +6
  
  so wee have the infinite solutions for which the roots are rational integers
  
  hence, the answer is infinite
• 10 years agoHelpfull: Yes(19) No(1)
• x^2-5x+c
  x=5+
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  c=0,x=5or0
  c=4,x=4or1
  c=6,x=3or2
  c=-6,x=6or-1
  c=-14,x=7or-2
  c=-24,x=8or-3
  this will satisfy for infinite value of c.
  hence,
  answer=infinite
  
  
  
  
  
• 12 years agoHelpfull: Yes(8) No(1)
• infinite.
  x=(5+(25-4c)^(1/2))/2
  x=(5-(25-4c)^(1/2))/2
  c=4,6,-6,-14,-24.....infinite
• 12 years agoHelpfull: Yes(6) No(2)
• since leading coefficient is one and coefficient of x is integer i.e -5 so for every integral value of 'c' for which equation has rational roots. then these rational roots must be integral.
  and c is of the form given by
  
  25 - 4c = (2*n +_ 1)^2
  
  c =( 25 - (2*n +_ 1)^2)/4
  
  n belongs to integer
  
  so there exists infinitely many c.
• 8 years agoHelpfull: Yes(0) No(1)
• its infinite
• 2 years agoHelpfull: Yes(0) No(0)