To find the roots of the quadratic equation [tex]\(x^2 + 2x - 5 = 0\)[/tex], we need to solve for [tex]\(x\)[/tex]. We can do this best by using the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the given quadratic equation [tex]\(x^2 + 2x - 5 = 0\)[/tex]:
- [tex]\(a = 1\)[/tex],
- [tex]\(b = 2\)[/tex],
- [tex]\(c = -5\)[/tex].
First, let's calculate the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = (2)^2 - 4(1)(-5) = 4 + 20 = 24
\][/tex]
Next, we use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm \sqrt{24}}{2(1)} = \frac{-2 \pm 2\sqrt{6}}{2}
\][/tex]
Simplifying the expression inside the fraction:
[tex]\[
x = \frac{-2 + 2\sqrt{6}}{2} \quad \text{and} \quad x = \frac{-2 - 2\sqrt{6}}{2}
\][/tex]
This gives us:
[tex]\[
x = -1 + \sqrt{6} \quad \text{and} \quad x = -1 - \sqrt{6}
\][/tex]
Thus, the two roots of the equation [tex]\(x^2 + 2x - 5 = 0\)[/tex] are:
[tex]\[
x = -1 + \sqrt{6} \quad (Option \, D) \quad \text{and} \quad x = -1 - \sqrt{6} \quad (Option \, B)
\][/tex]
So the correct choices are:
[tex]\[
B. \quad x = -1 - \sqrt{6}
\][/tex]
[tex]\[
D. \quad x = -1 + \sqrt{6}
\][/tex]
