Certainly! Let's solve the quadratic equation [tex]\(x^2 + 6x + 1 = 0\)[/tex].
### Step 1: Identify the coefficients
A quadratic equation is generally written as [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, we have:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = 1\)[/tex]
### Step 2: Calculate the discriminant (D)
The discriminant [tex]\(D\)[/tex] is given by the formula:
[tex]\[ D = b^2 - 4ac \][/tex]
Substituting in our coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ D = 6^2 - 4 \cdot 1 \cdot 1 \][/tex]
[tex]\[ D = 36 - 4 \][/tex]
[tex]\[ D = 32 \][/tex]
### Step 3: Calculate the roots
The roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{D}}}{{2a}} \][/tex]
Given [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(D = 32\)[/tex], we can find the two roots.
#### Root 1:
[tex]\[ x_1 = \frac{{-b + \sqrt{D}}}{{2a}} \][/tex]
[tex]\[ x_1 = \frac{{-6 + \sqrt{32}}}{{2 \cdot 1}} \][/tex]
[tex]\[ x_1 = \frac{{-6 + \sqrt{32}}}{{2}} \][/tex]
#### Root 2:
[tex]\[ x_2 = \frac{{-b - \sqrt{D}}}{{2a}} \][/tex]
[tex]\[ x_2 = \frac{{-6 - \sqrt{32}}}{{2 \cdot 1}} \][/tex]
[tex]\[ x_2 = \frac{{-6 - \sqrt{32}}}{{2}} \][/tex]
Calculating these, we get:
- Root 1 ([tex]\(x_1\)[/tex]) ≈ -0.1715728752538097
- Root 2 ([tex]\(x_2\)[/tex]) ≈ -5.82842712474619
Thus, the roots of the equation are:
[tex]\[ (-0.1715728752538097, -5.82842712474619) \][/tex]
### Step 4: Determine the number of roots (N)
A quadratic equation always has 2 solutions or roots (which may be real or complex). In this case, since the discriminant [tex]\(D = 32\)[/tex] is positive, the equation has two distinct real roots.
Therefore, the number of roots [tex]\(N\)[/tex] is:
[tex]\[ N = 2 \][/tex]
### Summary
- Discriminant (D) = 32
- Roots = (-0.1715728752538097, -5.82842712474619)
- Number of roots (N) = 2
So, the solution to the given quadratic equation [tex]\(x^2 + 6x + 1 = 0\)[/tex] is fully described by these values.
