Select the two values of [tex]\(x\)[/tex] that are roots of this equation:

[tex]\[x^2 + 2x - 5 = 0\][/tex]

A. [tex]\(x = -1 + 2\sqrt{6}\)[/tex]

B. [tex]\(x = -1 - \sqrt{6}\)[/tex]

C. [tex]\(x = -1 + \sqrt{6}\)[/tex]

D. [tex]\(x = -1 - 2\sqrt{6}\)[/tex]



Answer :

To find the roots of the equation [tex]\( x^2 + 2x - 5 = 0 \)[/tex], we need to solve it step-by-step. We will use the quadratic formula:

[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]

In this quadratic equation, the coefficients are:
[tex]\[ a = 1, \quad b = 2, \quad c = -5 \][/tex]

First, we calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-5) \][/tex]
[tex]\[ \Delta = 4 + 20 \][/tex]
[tex]\[ \Delta = 24 \][/tex]

Now, we will use the discriminant to find the roots using the quadratic formula:
[tex]\[ x = \frac{{-2 \pm \sqrt{24}}}{2 \cdot 1} \][/tex]

We know that [tex]\(\sqrt{24} = 2\sqrt{6}\)[/tex]. Therefore, substituting it back into the equation gives:
[tex]\[ x = \frac{{-2 \pm 2\sqrt{6}}}{2} \][/tex]

We can simplify the expression:
[tex]\[ x = \frac{{-2}}{2} \pm \frac{{2\sqrt{6}}}{2} \][/tex]
[tex]\[ x = -1 \pm \sqrt{6} \][/tex]

Thus, the roots of the equation [tex]\( x^2 + 2x - 5 = 0 \)[/tex] are:
[tex]\[ x = -1 + \sqrt{6} \quad \text{and} \quad x = -1 - \sqrt{6} \][/tex]

So, the correct answers are:
C. [tex]\(x = -1 + \sqrt{6}\)[/tex]
B. [tex]\(x = -1 - \sqrt{6}\)[/tex]