how many values of c in x^2+5x+c,result in rational roots which are integers?

• yippe..i found out the soln..check out below:
  
  
  
  There are infinitely many such c.
  
  By the quadratic formula, the roots of x2−5x+c=0 are
  
  x=5±25−4c√2
  
  In order for these roots to be rational the discriminant 25−4c needs to be a perfect square, and in order for the roots to be integers the discriminant needs to be the square of an odd number. So in order to have integer roots we must have
  
  25−4c=(2k+1)2
  
  for some integer k, and therefore
  
  c=25−(2k+1)24
  
  As k ranges over all integers it's not to hard to see that the range of c is infinite.
• 9 years agoHelpfull: Yes(4) No(3)
• c= 0,4,6,-6,-14.........................................infinite
• 9 years agoHelpfull: Yes(3) No(0)
• I think there will be 4 solution......0,4,6,-6 bcz......
  from the equation we will get x= x=5±(25−4c)^(1/2)/2.........so to make this integer the inside thing has to be perfect sqauare......so on putting 1 we 0...we will get 25 so it is a perfect square also putting 4 we got 16....and putting 6 we got 1....and on putting -6 we got 49...so all are perfect square ...so there will be 4 solutions of c.....
• 9 years agoHelpfull: Yes(1) No(2)
• 3 values
  c=0,4,6
• 9 years agoHelpfull: Yes(0) No(3)
• the roots will be rational and real only when discriminant D is positive and a perfect square
  here d=5^2-4*c
  d will be positive and real for only two values of c which are c=1 and 4.
  so answer is 2
• 9 years agoHelpfull: Yes(0) No(1)
• @anjali
  x=(-b+- sq root(b2 -4ac))/ 2a
  the value of x,you got is not crrct..
• 9 years agoHelpfull: Yes(0) No(3)
• Ans will be 0 ,4 and 6
  first will gives b^2-4ac>0
  second root must be int.
  their are only three value of cwhere dis satisfyd at 0,4 and 6.
  
• 9 years agoHelpfull: Yes(0) No(0)