How many roots does y=3x^2+8 have?

This quadratic has two unequal, imaginary roots.

Step-by-step explanation:

y = 3x^2 + 8 is a quadratic function, and (like all quadratics) it has two roots.  In this case they happen to be different (unequal), as well as imaginary.

Setting y = 3x^2 + 8 = 0, we get:

3x^2 = -8, or

x^2 = -8/3

Taking the square root of both sides, we get

x = ±i√(8/3) (2 imaginary roots)

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point