Démontrer la formule de la forme canonique d'un polynôme du second degré Problème

f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})

f(x)=a(x^2+\frac{b}{a}x+\frac{c}{b})

f(x)=a(x^2+bx+c)

f(x)=a(x^2+\frac{a}{b}x+\frac{a}{c})

(x+\frac{b}{2a})^2 = x^2+\frac{b}{a}x+\frac{b^2}{4a^2}

(x+\frac{b}{2a})^2 = x^2+\frac{b^2}{4a^2}

(x+\frac{b}{2a})^2 = x^2+\frac{b}{a}x

(x+\frac{b}{2a})^2 = x^2+\frac{b}{a}x+\frac{b^2}{2a^2}

f(x) = a \left ((x+\frac{b}{2a})^2- \frac{b^2}{4a^2}+\frac{c}{a} \right )

f(x) = a \left ((x+\frac{b}{2a})^2+\frac{c}{a} \right )

f(x) = a \left ((x+\frac{b}{2a})^2- \frac{b}{2a}+\frac{c}{a} \right )

f(x) = a \left ((x+\frac{b}{2a})^2- \frac{b^2}{4a^2}+\frac{c^2}{a^2} \right )

\alpha = \dfrac{-b}{2a}
\beta = -\dfrac{b^2-4ac}{4a}

\alpha = \dfrac{b}{2a}
 \beta =\dfrac{b^2-4ac}{4a}

\alpha = \dfrac{-b}{a}
\beta = -\dfrac{b^2-4ac}{a}

\alpha = -\dfrac{b^2-4ac}{4a}
\beta =\dfrac{-b}{2a}