Odpowiedź:
zad. 6
[tex]y = - 2(x - 4)(x + 5) \\ y = - 2( {x}^{2} + 5x - 4x - 20) \\ y = - 2 {x}^{2} - 2x + 40[/tex]
równanie osi symetrii
[tex]x = \frac{ - b}{2a} [/tex]
[tex]x = \frac{ - ( - 2)}{2 \times ( - 2)} = \frac{2}{ - 4} = - \frac{1}{2} [/tex]
współrzędne wierzchołka
W(p, q)
[tex]p = \frac{ - b}{2a} = \frac{ - ( - 2)}{2 \times ( - 2)} = \frac{2}{ - 4} = - \frac{1}{2} [/tex]
[tex]∆ = ( - 2 {)}^{2} - 4 \times ( - 2) \times 40 = 4 + 320 = 324[/tex]
[tex]q = - \frac{∆}{4a} = - \frac{ 324}{4 \times ( - 2)} = - \frac{324}{ - 8} = 40.5[/tex]
zad. 7
[tex] {x}^{4} - 12 {x}^{2} + 27 = 0 \\ {x}^{2} = t \\ {t}^{2} - 12t + 27 = 0 \\ ∆ = ( - 12 {)}^{2} - 4 \times 1 \times 27 = 144 - 108 = 36 \\ \sqrt{∆} = \sqrt{36} = 6 \\ t_{1} = \frac{ - b - \sqrt{∆} }{2a} = \frac{ - ( - 12) - 6}{2 \times 1} = \frac{12 - 6}{2} = \frac{6}{2} = 3 \\ t_{2} = \frac{ - b + \sqrt{∆} }{2a} = \frac{ 12 + 6}{2} = \frac{18}{2} = 9 \\ x_{1} = \sqrt{3} \: \: \: \: x_{2} = - \sqrt{3} \: \: \: \: x_{3} = 3 \: \: \: \: x_{4} = - 3[/tex]
