What kind of solutions does #3z^2 + z - 1 = 0 # have?

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Answer 1 · 2015-07-05T16:45:51

The discriminant (the thing we take the square root of in the quadratic formula) is:

#b^2 -4ac#.

In #3z^2 + z - 1 = 0 #, we have

#a = 3#
#b = 1#
#c = -1#

So
#b^2 -4ac = (1)^2 - 4 (3)(-1) = 1+12 = 13#.

#13# is positive, so there are two distinct real solutions. It is not a perfect square, so the solutions are irrational.

The equation has two distinct irrational real solutions.