Question 100882
D) Rule: If (b^2-4ac) >= 0 then you have a real number solution.
A)-4ac=-4*3*-3=36 so b can be any real number.
B)-4ac=-4*5*1=-20 so if b<= -sqrt(20)= -2sqrt(5) or b>= 2sqrt(5) there will be a real number solution.
C)-4ac=-4*-3*-3=-36 so if b<=-6 or b>=6 there will be a real number solution.
Ed