Równanie kwadratowe: Różnice pomiędzy wersjami

Ponieważ

: <math>\begin{align} ax^2 + bx + c & = a\left(x^2 + \tfrac{bx}{a} + \tfrac{c}{a}\right) = \\ & = a\left(x^2 + \tfrac{xb}{a} + \tfrac{4ac}{4a^2}\right) = \\ & = a\left(x^2 + \tfrac{2xb}{2a} + \tfrac{4ac - b^2}{4a^2} + \tfrac{b^2}{4a^2}\right) = \\ & = a\left(x^2 + \tfrac{2xb}{2a} + \tfrac{b^2}{4a^2} - \tfrac{b^2 - 4ac}{4a^2}\right) = \\ & = a\left((x + \tfrac{b}{2a})^2 - \tfrac{b^2 - 4ac}{4a^2}\right) = \\ & = a\left(x + \tfrac{b}{2a} - \tfrac{\sqrt{b^2 - 4ac}}{2a}\right)\left(x + \tfrac{b}{2a} + \tfrac{\sqrt{b^2 - 4ac}}{2a}\right) = \\ & = a\left(x - \tfrac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \tfrac{-b - \sqrt{b^2 - 4ac}}{2a}\right) \end{align}</math>

oraz

: <math>x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}.</math>

/><span style="color:#bc1e47">■</span> =0: −⁴⁄₃''x''²+⁴⁄₃''x''−¹⁄₃<br

/><span style="color:#0081cd">■</span> &gt;0: ³⁄₂''x''²+¹⁄₂''x''−⁴⁄₃]]

 

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