how many values of c in the equation x 2-5x c 0 results in rational roots which are integer a

roots will be x=(5+-sqrt(25-c))/2
now,for it to be an integer sqrt value must be odd as there is a 2 in divide.so we will look for those values of c which makes it a odd.values of c are 0,4,6..also we can have negative values...so options a,b,c are invalid.infinite is the answer
10 years agoHelpfull: Yes(1) No(0)
it has infinite solutions some of the value of c are 6,4,-6,-14 etc...
10 years agoHelpfull: Yes(0) No(4)
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...
Hence, Infinite is the answer
10 years agoHelpfull: Yes(0) No(2)
ans-a
b^2-4ac>0
5^2-4*1*c>0
25-4c>=0
for c=4
9>0
so ans is 1 value of c is 4
10 years agoHelpfull: Yes(0) No(4)
comparing with x^2 - Sx + P=0;
S=5;
hence infinite pair of integers with sum=5;
10 years agoHelpfull: Yes(0) No(0)
x^2-5x+c=0
=> x= [10+-sqrt(100-16c)]/4

for, c = 0,4,6,-6,-14,-24,... we get rational roots which are integers
so,c has infinite no. of values.
d) infinite
10 years agoHelpfull: Yes(0) No(0)

how many values of c in the equation
x^2-5x+c=0, results in rational roots which are integer.
a) 1
b) 3
c) 0
d) infinite