Hello from MrBillDoesMath!
Answer:
All equations have two real roots.
Discussion:
The solutions of the quadratic ax^2 + bx + c = 0 are given by the formula:
x = (-b +\- sqrt(b^2-4ac) ) / (2a)
The key element is the discriminant, D, which equals (b^2-4ac)
For 2) a = 3, b = 4, c = -4, D = 4^2 - 4(3)(-4) = 16 + 48 = 64 > 0 so there are two real solutions
For 4) D = b^2 - 4ac = (-1)^2 - 4(6)(-1) = 1 + 24 = 25 >0 so there are two real solutions
For 6), a = -3, b = 12, c= -11, D = 12^2 - 4(-3)(-11) = 144 - 132 = 12 > 0 so
there are two real solutions
Thank you,
MrB
