How many values of C in the equation x2 - 5x + c result in rational roots which are integers?

A. 1

B. 3

C. 6

D. Infinite

• Option d. Reason: if we have quadratic equation of form x^2 - 5x + c = 0, then solution for roots will be x = (1/2a) * (-b +/- sqrt(b^2 - 4ac)) [formula]
  or, x = (1/2) * (5 +/- sqrt(25 - 4c))
  As, it may be observed, if sqrt() term is perfect square then it would be rational root AND for being an integer root the numerator should be even i.e. sqrt() term should be odd. By keeping this fact in mind, we have 3 such values for c>=0 namely 0,4 and 6. Other values will yield imaginary roots.
  Since there is no restriction on c value 'c' can also be a -ve number(c
• 9 years agoHelpfull: Yes(9) No(0)
• There are infinitely many such c.
  
  By the quadratic formula, the roots of x^2-5x+c=0 are
  
  x= frac{5pmsqrt{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 = frac{25-(2k+1)^2}{4}
  
• 9 years agoHelpfull: Yes(3) No(0)
• only 1 value that is 6.
• 9 years agoHelpfull: Yes(0) No(5)
• There are infinitely many such c.
  
  By the quadratic formula, the roots of x^2-5x+c=0 are
  
  x= 2{25-4c}/{2}
  
  so there are no. of possible outcome
• 9 years agoHelpfull: Yes(0) No(0)
• ANS : infinite
  
  sloution:The value of x can be calculated as:
  
  Formula : -b+-rootb^2-4ac/2a
  
  x=( 5 +- sqrt(5^2 -4c))/2
  
  For c = 0 , 4 , 6 , -6, -14, -24 , -36, -50 ……….and so on we can get rational roots which are
  Integers.
  Hence the Answer is D: Infinite
  
  
• 7 years agoHelpfull: Yes(0) No(0)