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.

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