Let's solve the given quadratic equation [tex]\(2x^2 + 6x + 1 = 0\)[/tex].
1. Identifying the coefficients:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 6 \)[/tex]
- [tex]\( c = 1 \)[/tex]
2. Calculate the discriminant (D):
The discriminant is calculated using the formula:
[tex]\[
D = b^2 - 4ac
\][/tex]
Substituting the values, we have:
[tex]\[
D = 6^2 - 4 \cdot 2 \cdot 1 = 36 - 8 = 28
\][/tex]
3. Determine the nature and number of roots based on the discriminant:
- If [tex]\( D > 0 \)[/tex], there are two distinct real roots.
- If [tex]\( D = 0 \)[/tex], there is exactly one real root (repeated).
- If [tex]\( D < 0 \)[/tex], there are no real roots; the roots are complex.
Since [tex]\( D = 28 \)[/tex] and [tex]\( D > 0 \)[/tex], we conclude that there are two distinct real roots.
4. Calculate the roots:
The roots of the quadratic equation are given by the formulas:
[tex]\[
x_1 = \frac{{-b + \sqrt{D}}}{{2a}}
\][/tex]
[tex]\[
x_2 = \frac{{-b - \sqrt{D}}}{{2a}}
\][/tex]
Substituting the values:
[tex]\[
x_1 = \frac{{-6 + \sqrt{28}}}{{2 \cdot 2}} = \frac{{-6 + \sqrt{28}}}{4}
\][/tex]
[tex]\[
x_2 = \frac{{-6 - \sqrt{28}}}{{2 \cdot 2}} = \frac{{-6 - \sqrt{28}}}{4}
\][/tex]
5. Simplify the numerical results:
- Approximation for [tex]\(\sqrt{28}\)[/tex]:
[tex]\[
\sqrt{28} \approx 5.291502622
\][/tex]
- Therefore, the roots are approximately:
[tex]\[
x_1 = \frac{{-6 + 5.291502622}}{4} \approx -0.17712434446770464
\][/tex]
[tex]\[
x_2 = \frac{{-6 - 5.291502622}}{4} \approx -2.8228756555322954
\][/tex]
6. Summary:
- Roots:
[tex]\[
x_1 \approx -0.17712434446770464
\][/tex]
[tex]\[
x_2 \approx -2.8228756555322954
\][/tex]
- Discriminant:
[tex]\[
D = 28
\][/tex]
- Number of roots (N):
[tex]\[
N = 2
\][/tex]
Putting it all together, we get:
[tex]\[
\boxed{
\begin{array}{l}
2x^2 + 6x + 1 = 0 \\
\text{Roots:} \; x_1 \approx -0.17712434446770464, \; x_2 \approx -2.8228756555322954 \\
D = 28 \\
N = 2 \\
\end{array}
}
\][/tex]
