WisFaq!

Formule omwerken

van ax²+bx+c naar a(x-$\alpha$)²+$\beta$

a(x²+(b/a)x+(c/a)
a(x²+(2b/2a)x+(b²/4a²)-(b²/4a²)+(c/a))
a(x+(b/2a))²-(b²/4a²)+(c/a)

hoe kan ik nu van -(b²/4a²)+(c/a) naar -b²+4ac/4a
zelfde noemer zetten maakt toch
=-(b²/4a²)+(4ac/4a²)
=(-b²+4ac)/4a²

en zo klopt mijn noemer niet....

Tim B.
31-8-2014

Antwoord

$
\begin{array}{l}
ax^2 + bx + c = \\
a\left( {x^2 + \frac{b}{a}x + \frac{c}{a}} \right) = \\
a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{c}{a}} \right) = \\
a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{c}{a} \cdot \frac{{4a}}{{4a}}} \right) = \\
a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{{4ac}}{{4a^2 }}} \right) = \\
a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a^2 }}} \right) = \\
a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a}} = \\
a\left( {x - \frac{{ - b}}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a}} \\
\alpha = \frac{{ - b}}{{2a}}\,\,en\,\,\beta = \frac{{ - b^2 + 4ac}}{{4a}} \\
\end{array}
$

Dus...

WvR
31-8-2014