AB: Gleichungen 1. bis 3. Grades

2·x + 4 = 0   | -4
2·x = -4   | :2
x = -2

10·x - 10 = 0   | +10
10·x = 10   | :10
x = 1

9·x - 3 = 0   | +3
9·x = 3   | :9
x = \( \frac{3}{9} \)   | kürzen mit :3
x = \( \frac{1}{3} \)

-4·x + 12 = 0 | -12
-4·x = -12 | :(-4)
x = 3

\( x^2 + x - 2 = 0 \\ x^2 + 1·x + (-2) = 0 \\ p = 1, q = -2 \\ x_{1,2} = -( \frac{p}{2} ) \pm \sqrt{ ( \frac{p}{2} )^2 - q} \\ x_{1,2} = -( \frac{1}{2} ) \pm \sqrt{ ( \frac{1}{2} )^2 - (-2) } \\ x_{1,2} = -\frac{1}{2} \pm \sqrt{ \frac{1}{4} + \frac{8}{4} } \\ x_{1,2} = -\frac{1}{2} \pm \sqrt{ \frac{9}{4} } \\ x_{1,2} = -\frac{1}{2} \pm \frac{3}{2} \\ x_1 = -\frac{1}{2} + \frac{3}{2} = 1 \\ x_2 = -\frac{1}{2} - \frac{3}{2} = -2 \)

\( x^2 + 4·x - 5 = 0 \\ x^2 + 4·x + (-5) = 0 \\ p = 4, q = -5 \\ x_{1,2} = -( \frac{p}{2} ) \pm \sqrt{ ( \frac{p}{2} )^2 - q} \\ x_{1,2} = -( \frac{4}{2} ) \pm \sqrt{ ( \frac{4}{2} )^2 - (-5) } \\ x_{1,2} = -2 \pm \sqrt{ ( 2 )^2 - (-5) } \\ x_{1,2} = -2 \pm \sqrt{9} \\ x_1 = -2 + 3 = 1 \\ x_2 = -2 - 3 = -5 \)

\( 2·x^2 - 2·x - 12 = 0 \\ a = 2, b = -2, c = -12 \\ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4·a·c}}{2·a} \\ x_{1,2} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4·2·(-12)}}{2·2} \\ x_{1,2} = \frac{2 \pm \sqrt{4 + 96}}{4} \\ x_{1,2} = \frac{2 \pm \sqrt{100}}{4} \\ x_{1,2} = \frac{2 \pm 10}{4} \\ x_1 = \frac{2 + 10}{4} = 3 \\ x_2 = \frac{2 - 10}{4} = -2 \)

\( 8·x^2 + 2x - 3 = 0 \\ a = 8, b = 2, c = -3 \\ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4·a·c}}{2·a} \\ x_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4·8·(-3)}}{2·8} \\ x_{1,2} = \frac{-2 \pm \sqrt{4 + 96}}{16} \\ x_{1,2} = \frac{-2 \pm \sqrt{100}}{16} \\ x_{1,2} = \frac{-2 \pm 10}{16} \\ x_1 = \frac{-2 + 10}{16} = \frac{8}{16} = 0,5 \\ x_2 = \frac{-2 - 10}{16} = \frac{-12}{16} = -0,75 \)

x³ - x = (x² - 1)·x → x₁ = 0

Mit der p-q-Formel (oder durch genaues Hinsehen) findet man nun die Nullstellen des Polynoms x² - 1 mit:

\( x_2 = +\sqrt{1} = 1 \\ x_3 = -\sqrt{1} = -1 \)

Alternativ funktioniert auch die Mitternachtsformel mit a = 1, b = 0 und c = -1.

x³ - x² - 6·x = (x² - x - 6)·x → x₁ = 0

Mit der p-q-Formel findet man die Nullstellen des Polynoms x² - x - 6:

\( x^2 - x - 6 = 0 \\ x^2 + (-1)·x + (-6) = 0 \\ p = -1 \text{ und } q = -6 \\ x_{2,3} = -( \frac{p}{2} ) \pm \sqrt{ ( \frac{p}{2} )^{2} - q} \\ x_{2,3} = -( \frac{-1}{2} ) \pm \sqrt{ ( \frac{-1}{2} )^{2} - (-6)} \\ x_{2,3} = \frac{1}{2} \pm \sqrt{ \frac{1}{4} + 6} \\ x_{2,3} = \frac{1}{2} \pm \sqrt{ \frac{25}{4} } \\ x_{2,3} = \frac{1}{2} \pm \frac{5}{2} \\ x_2 = \frac{1}{2} + \frac{5}{2} = \frac{6}{2} = 3 \\ x_3 = \frac{1}{2} - \frac{5}{2} = -\frac{4}{2} = -2 \)

Alternativ funktioniert auch hier die Mitternachtsformel mit a = 1, b = -1 und c = -6.