Question

Which are the solutions of the quadratic equation?

x2 = –5x – 3

–5, 0
StartFraction negative 5 minus StartRoot 13 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 13 EndRoot Over 2 EndFraction
StartFraction 5 minus StartRoot 13 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 13 EndRoot Over 2 EndFraction
5, 0

Answer 1

To solve the quadratic equation \(x^2 = -5x - 3\), we can rewrite it in standard form:

\[
x^2 + 5x + 3 = 0
\]

Now we can use the quadratic formula, which is given by:

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

Here, \(a = 1\), \(b = 5\), and \(c = 3\).

First, we need to calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 3 = 25 - 12 = 13
\]

Now we can substitute back into the quadratic formula:

\[
x = \frac{-5 \pm \sqrt{13}}{2 \cdot 1} = \frac{-5 \pm \sqrt{13}}{2}
\]

This gives us the two solutions:

1. \(x = \frac{-5 - \sqrt{13}}{2}\)
2. \(x = \frac{-5 + \sqrt{13}}{2}\)

Among the options provided, the correct ones are:

\[
\frac{-5 - \sqrt{13}}{2}, \frac{-5 + \sqrt{13}}{2}
\]

So, the solutions of the quadratic equation \(x^2 = -5x - 3\) are:

\[
\text{StartFraction negative 5 minus StartRoot 13 EndRoot Over 2 EndFraction, StartFraction negative 5 plus StartRoot 13 EndRoot Over 2 EndFraction}
\]