To determine which two values of [tex]\( x \)[/tex] are the roots of the polynomial [tex]\( x^2 + 3x + 5 \)[/tex], let's solve the quadratic equation using the quadratic formula. The quadratic formula for the equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In the polynomial [tex]\( x^2 + 3x + 5 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 5 \)[/tex]
First, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 5 = 9 - 20 = -11
\][/tex]
Since the discriminant is negative ([tex]\( \Delta = -11 \)[/tex]), the solutions will be complex numbers. Now, substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( \Delta \)[/tex] back into the quadratic formula:
[tex]\[
x = \frac{-3 \pm \sqrt{-11}}{2 \cdot 1} = \frac{-3 \pm \sqrt{-11}}{2}
\][/tex]
The roots are:
[tex]\[
x = \frac{-3 + \sqrt{-11}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{-11}}{2}
\][/tex]
Next, let's match these roots with the provided options:
- [tex]\( A. \ x = \frac{-3 - \sqrt{11}}{2} \)[/tex]
- [tex]\( B. \ x = \frac{-3 - \sqrt{-11}}{2} \)[/tex]
- [tex]\( C. \ x = \frac{-3 + \sqrt{11}}{2} \)[/tex]
- [tex]\( D. \ x = \frac{-3 - \sqrt{2}}{2} \)[/tex]
- [tex]\( E. \ x = \frac{-3 + \sqrt{29}}{2} \)[/tex]
- [tex]\( F. \ x = \frac{-3 + \sqrt{-11}}{2} \)[/tex]
From the solutions, we see that the correct values are:
- [tex]\( \frac{-3 - \sqrt{-11}}{2} \)[/tex] matches option [tex]\( B \)[/tex]
- [tex]\( \frac{-3 + \sqrt{-11}}{2} \)[/tex] matches option [tex]\( F \)[/tex]
Therefore, the correct values of [tex]\( x \)[/tex] that are roots of the polynomial [tex]\( x^2 + 3x + 5 \)[/tex] are:
- [tex]\( B. \ x = \frac{-3 - \sqrt{-11}}{2} \)[/tex]
- [tex]\( F. \ x = \frac{-3 + \sqrt{-11}}{2} \)[/tex]
Thus, the two values of [tex]\( x \)[/tex] that are roots of the polynomial are [tex]\( \boxed{B \text{ and } F} \)[/tex].
