Of course! Let's solve the quadratic equation [tex]\(x^2 + 6x + 8 = 0\)[/tex] using the quadratic formula. The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = 8\)[/tex]
Now follow these steps:
1. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 6^2 - 4 \cdot 1 \cdot 8
\][/tex]
[tex]\[
\Delta = 36 - 32
\][/tex]
[tex]\[
\Delta = 4
\][/tex]
2. Find the two solutions using the quadratic formula:
Since the discriminant ([tex]\(\Delta\)[/tex]) is 4, we have two real solutions. We use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
First solution ([tex]\(x_1\)[/tex]):
[tex]\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a}
\][/tex]
Substitute:
[tex]\[
x_1 = \frac{-6 + \sqrt{4}}{2 \times 1}
\][/tex]
[tex]\[
x_1 = \frac{-6 + 2}{2}
\][/tex]
[tex]\[
x_1 = \frac{-4}{2}
\][/tex]
[tex]\[
x_1 = -2.0
\][/tex]
Second solution ([tex]\(x_2\)[/tex]):
[tex]\[
x_2 = \frac{-b - \sqrt{\Delta}}{2a}
\][/tex]
Substitute:
[tex]\[
x_2 = \frac{-6 - \sqrt{4}}{2 \times 1}
\][/tex]
[tex]\[
x_2 = \frac{-6 - 2}{2}
\][/tex]
[tex]\[
x_2 = \frac{-8}{2}
\][/tex]
[tex]\[
x_2 = -4.0
\][/tex]
Therefore, the solutions to the quadratic equation [tex]\(x^2 + 6x + 8 = 0\)[/tex] are:
[tex]\[
x_1 = -2.0
\][/tex]
[tex]\[
x_2 = -4.0
\][/tex]
The discriminant is 4, and the solutions are [tex]\(x_1 = -2.0\)[/tex] and [tex]\(x_2 = -4.0\)[/tex].
