If a quadratic equation with real coefficients has a discriminant of 3 then the two roots must be

In a quadratic equation

q(x) = ax^2 + bx + c 

The discriminant is = b^2 - 4ac

We have that discriminant = 3

If b^2 - 4ac > 0, then the roots are real.

If b^2 - 4ac < 0 then the roots are imaginary

In this problem b^2 - 4ac > 0   3 > 0

then the two roots must be real

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

thats not a system of equations, that is a solution