Hi guys here is a similar problem to the one I posted yesterday. I tried applying the b^2-4ac equation didnt work out as planned. Please help thanks
How many integers can let x^2+bx+c=0? B and c are constants.
2) b= - (c+1) Dont know why the answer is E
b^2-4ac = 0 means that it has only one real root.
b^2-4ac = m^2 (where m is integer) means that it has two real roots, but it can be an interger or a rational number
S1. c<0 tells b^2-4c>0 thus, two real roots
x= [-b+-sqrt(b^2-4c)]/2 we don't know x value will be an integer or just areal number. insuff.
S2. b = -(c+1)
x = [c+1+-|c-1|]/2 still insuff. it can be an interger or a rational number
X= c or 1
Combine S1 and S2.. insuff.
X^2-(c+1)X +c = 0
X = [c+1+(1-c)]/2 or [c+1-(1-c)]/2
X = 1 or c (c is just a constant we don't know it's integer or not)
The only thing that matters is what you believe.