Question

1. solve $x^2 + 5x = 3$.

a. $x = \frac{-5 + \sqrt{37}}{2}$ and $x = \frac{-5 - \sqrt{37}}{2}$
b. $x = \frac{5 + \sqrt{37}}{2}$ and $x = \frac{5 - \sqrt{37}}{2}$
c. $x = \frac{5 + \sqrt{13}}{2}$ and $x = \frac{5 - \sqrt{13}}{2}$
d. $x = -\frac{5 + \sqrt{13}}{2}$ and $x = -\frac{5 - \sqrt{13}}{2}$

Explanation:

Step1: Rewrite the equation in standard form

The given equation is \(x^{2}+5x = 3\). We rewrite it in the standard quadratic form \(ax^{2}+bx + c=0\) by subtracting 3 from both sides:
\(x^{2}+5x - 3=0\)
Here, \(a = 1\), \(b = 5\), and \(c=- 3\).

Step2: Apply the quadratic formula

The quadratic formula for solving \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 1\), \(b = 5\), and \(c=-3\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(5)^{2}-4\times1\times(-3)=25 + 12=37\)
Then, \(x=\frac{-5\pm\sqrt{37}}{2}\), which means \(x=\frac{-5+\sqrt{37}}{2}\) and \(x=\frac{-5 - \sqrt{37}}{2}\)

Answer:

A. \(x=\frac{-5+\sqrt{37}}{2}\) and \(x=\frac{-5-\sqrt{37}}{2}\)