\require{AMSmath} 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/4azelfde noemer zetten maakt toch=-(b2/4a2)+(4ac/4a2)=(-b2+4ac)/4a2en zo klopt mijn noemer niet.... Tim B. 2de graad ASO - zondag 31 augustus 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 zondag 31 augustus 2014 Re: Formule omwerken ©2001-2024 WisFaq
\require{AMSmath}
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/4azelfde noemer zetten maakt toch=-(b2/4a2)+(4ac/4a2)=(-b2+4ac)/4a2en zo klopt mijn noemer niet.... Tim B. 2de graad ASO - zondag 31 augustus 2014
Tim B. 2de graad ASO - zondag 31 augustus 2014
$\begin{array}{l} ax^2 + bx + c = \\ a\left( {x^2 + \frac{b}{a}x + \frac{c}{a}} \right) = \\ a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{c}{a}} \right) = \\ a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{c}{a} \cdot \frac{{4a}}{{4a}}} \right) = \\ a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 - \frac{{b^2 }}{{4a^2 }} + \frac{{4ac}}{{4a^2 }}} \right) = \\ a\left( {\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a^2 }}} \right) = \\ a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a}} = \\ a\left( {x - \frac{{ - b}}{{2a}}} \right)^2 + \frac{{ - b^2 + 4ac}}{{4a}} \\ \alpha = \frac{{ - b}}{{2a}}\,\,en\,\,\beta = \frac{{ - b^2 + 4ac}}{{4a}} \\ \end{array}$Dus... WvR zondag 31 augustus 2014
WvR zondag 31 augustus 2014
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