Question #8d622

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Answer 1 · 2017-05-08T10:57:04

$0$ $0$

We have:

$f (x) = 3 x^{2} + 3 x + 2$ $f(x) = 3x^2+3x+2$

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

$\Rightarrow 3 x^{2} + 3 x + 2 = 0$ $=> 3x^2+3x+2 = 0$

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]}