Question

1. solve the quadratic equation by factoring or by using the quadratic formula:

$3x^2 + 13x = -12$

$x = \underline{\quad\quad}$ and $\underline{\quad\quad}$

Explanation:

Step1: Rewrite in standard form

First, we rewrite the quadratic equation \(3x^{2}+13x = - 12\) in standard form \(ax^{2}+bx + c=0\) by adding 12 to both sides. So we get \(3x^{2}+13x + 12=0\). Here, \(a = 3\), \(b=13\), and \(c = 12\).

Step2: Use quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 3\), \(b = 13\), and \(c=12\) into the formula. First, calculate the discriminant \(\Delta=b^{2}-4ac=(13)^{2}-4\times3\times12=169 - 144 = 25\). Then, \(x=\frac{-13\pm\sqrt{25}}{2\times3}=\frac{-13\pm5}{6}\).

Step3: Find two solutions

For the plus sign: \(x=\frac{-13 + 5}{6}=\frac{-8}{6}=-\frac{4}{3}\). For the minus sign: \(x=\frac{-13-5}{6}=\frac{-18}{6}=- 3\). We can also factor the quadratic: \(3x^{2}+13x + 12=(3x + 4)(x+3)=0\). Setting each factor equal to zero gives \(3x+4 = 0\Rightarrow x=-\frac{4}{3}\) and \(x + 3=0\Rightarrow x=-3\).

Answer:

\(x=-\frac{4}{3}\) and \(x = - 3\) (or in the blanks: \(-\frac{4}{3}\) and \(-3\))