Math Solve equations Quadratic formula I & II

Quadratic equations are equations of the form:

$ax^2+bx+c=0$
i

### Hint

You can either use the pq-formula (Quadratic formula I) or the abc-formula (Quadratic formula II). Ultimately, however, you get the same result with both formulas.

## abc-formula

Given is a quadratic equation in the form: $\color{red}{a}x^2+\color{green}{b}x+\color{blue}{c}=0$.
The Quadratic formula II (abc-formula) for solving this equation is:

$x_{1,2} = \frac{-\color{green}{b} \pm \sqrt{\color{green}{b}^2 - 4\color{red}{a}\color{blue}{c}}}{2\color{red}{a}}$

### Example

Solve quadratic equation: $\color{red}{3}x^2+\color{green}{18}x+\color{blue}{15}=0$

1. #### Insert and simplify

$x_{1,2} = \frac{-\color{green}{b} \pm \sqrt{\color{green}{b}^2 - 4\color{red}{a}\color{blue}{c}}}{2\color{red}{a}}$

$x_{1,2} = \frac{-\color{green}{18} \pm \sqrt{\color{green}{18}^2 - 4\cdot\color{red}{3}\cdot\color{blue}{15}}}{2\cdot\color{red}{3}}$

$x_{1,2} = \frac{-18 \pm \sqrt{324 - 180}}{6}$

$x_{1,2} = \frac{-18 \pm \sqrt{144}}{6}$

$x_{1,2} = \frac{-18 \pm 12}{6}$
2. #### Calculate solutions

$x_{1} = \frac{-18 + 12}{6} = \frac{-6}{6}=-1$
$x_{2} = \frac{-18 - 12}{6} = \frac{-30}{6}=-5$

## pq-formula

Given is a quadratic equation in the canonical form: $x^2+\color{green}{p}x+\color{blue}{q}=0$.
The Quadratic formula I (pq-formula) for solving this equation is:

$x_{1,2} = -\frac{\color{green}{p}}{2} \pm\sqrt{(\frac{\color{green}{p}}{2})^2-\color{blue}{q}}$
!

### Remember

The pq-formula works only for quadratic equations in the canonical form.

### Example

Solve quadratic equation: $3x^2+18x+15=0$

1. #### Determine canonical form

$\color{red}{3}x^2+18x+15=0\quad|:\color{red}{3}$

$x^2+\color{green}{6}x+\color{blue}{5}=0$
2. #### Insert and simplify

$x_{1,2} = -\frac{\color{green}{p}}{2} \pm\sqrt{(\frac{\color{green}{p}}{2})^2-\color{blue}{q}}$

$x_{1,2} = -\frac{\color{green}{6}}{2} \pm\sqrt{(\frac{\color{green}{6}}{2})^2-\color{blue}{5}}$
$x_{1,2} = -3 \pm\sqrt{3^2-5}$
$x_{1,2} = -3 \pm\sqrt{4}$
$x_{1,2} = -3 \pm2$
3. #### Calculate solutions

$x_{1} = -3+2=-1$
$x_{2} = -3-2=-5$