Question

Question #8d622

Answer 1

`0`

Explanation:

We have:

`f(x) = 3x^2+3x+2`

An `x`-intercept occurs when `f(x)=0`

`=> 3x^2+3x+2 = 0`

To solve this quadratic we could compete the square or use the quadratic formula,

We can also use the discriminant `Delta=b^2-4ac` to determine the nature of the roots of `ax^2+bx+c=0`, as:

If `Delta=b^2-4ac \ { (lt 0, "no real roots"), (=0, "two equal real roots"), (gt 0, "two distinct real roots") :}`

For our equation we have:

`Delta = 3*3-4*2*2 < 0`

Hence there are no real solutions, and therefore `0` `x`-intercepts.

We can confirm this graphically:
graph{3x^2+3x+2 [-5, 5, -2, 5]}