Is there a solution for 3z^2 + z-1=0? And why?

Answer 1

there is one since the coefficients ofv z^2is positive and that -1 is hegative

the roots z=(-1+/-(1+12)^0.5)/6

z=0.434 and -0.767

Answer 2

there is a formula to solve quadratic equations.

for any equation ax^2+bx+c

D= b^2-4ac

and the factors of the equation is given by [-b±sqrt(D)] /2a

in this case the D = 1+12 = 13

and the factors are ( -1+√13)/6 and (-1-√13)/6

This is the solution because if you substitute z with either of these values you should get 0. That verifies that these are the values of z.

Answer 3

if you use the quadratic formula you get
.5(-1 +/- sqrt(1 + 12)) =
.5( -1 +/- sqrt(13))