Odpowiedź:
zad. 1
[tex]a)3 {x}^{2} + x + 1 = 0 \\ ∆ = {1}^{2} - 4 \times 3 \times 1 = 1 - 12 = - 11[/tex]
[tex]∆ < 0[/tex]
brak miejsc zerowych
[tex]b)2 {x}^{2} - 6x = 0 \\ ∆ = {( - 6)}^{2} - 4 \times 2 \times 0 = 36 - 0 = 36 \\ \sqrt{∆} = \sqrt{36} = 6 \\ x_{1} = \frac{ 6 - 6}{4} = 0 \\ x_{2} = \frac{6 + 6}{4} = \frac{12}{4} = 3[/tex]
[tex]c)5 {x}^{2} + 3x - 2 = 0 \\ ∆ = {3}^{2} - 4 \times 5 \times ( - 2) = 9 + 40 = 49 \\ \sqrt{∆} = \sqrt{49} = 7 \\ x_{1} = \frac{ - 3 - 7}{10} = \frac{ - 10}{10} = - 1 \\ x_{2} = \frac{ - 3 + 7}{10} = \frac{4}{10} = \frac{2}{5} [/tex]
zad. 2
[tex]y = {x}^{2} + 6x + 8 \\ ∆ = {6}^{2} - 4 \times 8 \times 1 = 36 - 32 = 4 \\ \sqrt{∆} = \sqrt{4} = 2 \\ x_{1} = \frac{ - 6 - 2}{2} = \frac{ -8 }{2} = - 4 \\ x_{2} = \frac{ - 6 + 2}{2} = \frac{ - 4}{2} = - 2 \\ p = - \frac{b}{2a} = - \frac{6}{2} = - 3 \\ q = - \frac{∆}{4a} = - \frac{2}{4} = - \frac{1}{2} \\ W( - 3; - \frac{1}{2} ) \\ f(0) = {0}^{2} + 6 \times 0 + 8 = 8[/tex]
