Real number solutions?

Question

Decide all of the values of b in the following equation that will give one or more real number solutions:
3x^2+bx-3=0

Answer 1

Delta = b^2 - 4ac has to be non negative

So:

Solve b^2 - 4 * 3 * (-3) >= 0

b^2 + 36 >= 0

b^2 >= -36

And this is so forall b, since b^2 >= 0

Ana

Answer 2

for this equation to have a real root that means the discriminant must be greater or equal to 0

so u solve b^2-4ac >=0

in this case:

b^2-(4)(3)(-3)>=0
b^2+36>=0
b^2>=-36

since b^2 is always >=0 therefore b can be any real number

Answer 3

sq. bothe area t^2 -13t + 37=a million t^2 -13t +36=0 use quadratic equation a=a million, b=-13 , c=36 t= (-b +_ sqr_/b^2-4ac)/2a t=(13+5)/2 =9 or t=(13-5)2 =4 then x^2=t (out of your given) x^2=9 , x=3 or x^2=4 , x=2