\require{AMSmath}

Dit is een reactie op vraag 88455

Re: Re: Berekenen van onbekende a en b

Sorry maar wat u doet snap ik, maar de waarde van a en b berekenen lukt me nog niet.

mboudd
Leerling mbo - dinsdag 17 september 2019

Antwoord

Met behulp van substitutie:

$
\begin{array}{l}
\left\{ \begin{array}{l}
- \frac{1}{2}a + b = 0 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
- \frac{1}{2} \cdot \left( {2\sqrt b } \right) + b = 0 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
- \sqrt b + b = 0 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b = \sqrt b \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b^2 = b \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b^2 - b = 0 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b\left( {b - 1} \right) = 0 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b = 0 \\
a = 2\sqrt b \\
\end{array} \right. \vee \left\{ \begin{array}{l}
b = 1 \\
a = 2\sqrt b \\
\end{array} \right. \\
\left\{ \begin{array}{l}
b = 0 \\
a = 0 \\
\end{array} \right.(v.n.) \vee \left\{ \begin{array}{l}
b = 1 \\
a = 2 \\
\end{array} \right. \\
\left\{ \begin{array}{l}
a = 2 \\
b = 1 \\
\end{array} \right. \\
\end{array}
$

Hoe moelijk kan dat zijn?

WvR
dinsdag 17 september 2019

Re: Re: Re: Berekenen van onbekende a en b

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