f(x) = ax² + bx i a ≠ 0 oraz W = (1; 3)
Δ = b² - 4 · a · 0 = b²
[tex]W = (\frac{-b}{2a}; \ \frac{-\Delta}{4a}) = (\frac{-b}{2a}; \ \frac{-b^2}{4a}) \ i \ W = (1; \ 3) \\ Zatem:\\ \frac{-b}{2a} = 1 \ i \ \frac{-b^2}{4a}= 3 \\\\ \frac{-b}{2a} = 1 \ \ \ |\cdot 2a \ i \ a \neq 0 \\ - b = 2a \ \ \ |\cdot (-1) \\ \underline{b = - 2a} \\\\ \frac{-b^2}{4a}= 3 \\ \frac{-(- 2a)^2}{4a}= 3 \\ \frac{-4a^2}{4a}= 3 \\ - a = 3 \ \ \ |\cdot (-1) \\ \underline{a = - 3} \\\\ b = - 2a \\ b = -2 \cdot (-3) \\ \underline{b = 6}[/tex]
Odp. A. a = - 3 i b = 6
