Question

$x^2 + 3x + 1 = 0$
what are the solutions to the equation?
$\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$
$\frac{-3 + \sqrt{13}}{2}$ and $\frac{-3 - \sqrt{13}}{2}$
$\frac{3 + \sqrt{5}}{3}$ and $\frac{3 - \sqrt{5}}{3}$
$\frac{-3 + \sqrt{5}}{2}$ and $\frac{-3 - \sqrt{5}}{2}$
$\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$

Explanation:

Step1: Identify coefficients for quadratic formula

For the quadratic equation \(ax^{2}+bx + c = 0\), here \(a = 1\), \(b = 3\), \(c = 1\).

Step2: Apply quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Substitute \(a = 1\), \(b = 3\), \(c = 1\) into the formula:
\(x=\frac{-3\pm\sqrt{3^{2}-4\times1\times1}}{2\times1}\)

Step3: Simplify the discriminant

Calculate \(3^{2}-4\times1\times1=9 - 4=5\).

Step4: Substitute discriminant back

So \(x=\frac{-3\pm\sqrt{5}}{2}\), which means the solutions are \(\frac{-3 + \sqrt{5}}{2}\) and \(\frac{-3-\sqrt{5}}{2}\).

Answer:

\(\frac{-3 + \sqrt{5}}{2}\) and \(\frac{-3-\sqrt{5}}{2}\) (corresponding to the option with these two solutions)