VVV#2

VVV#2

by Marna Zuhlsdorff -
Number of replies: 0
Gebruik vierkantsvoltooiing en bepaal  vir x: \( {x^2} - px - 4 = 0 \)


OPSIE 1

\( \begin{align} {x^2} - px - 4 &= 0\\ {x^2} - px &= 4\\ {x^2} - px + {\left( {\frac{p}{2}} \right)^2} &= 4 + {\left( {\frac{p}{2}} \right)^2}\\ {\left( {x - \frac{p}{2}} \right)^2} &= \frac{4}{1} + \frac{{{p^2}}}{4}\\ {\left( {x - \frac{p}{2}} \right)^2} &= \frac{{16 + {p^2}}}{4}\\ x - \frac{p}{2} &= \pm \sqrt {\frac{{16 + {p^2}}}{4}} \\ x - \frac{p}{2} &= \pm \frac{{\sqrt {{p^2} + 16} }}{{\sqrt 4 }}\\ x - \frac{p}{2} &= \pm \frac{{\sqrt {{p^2} + 16} }}{2}\\ x &= \frac{{p \pm \sqrt {{p^2} + 16} }}{2} \end{align} \)

OPSIE 2

\( \begin{align} {x^2} - px - 4 &= 0\\ {x^2} - px + {\left( {\frac{p}{2}} \right)^2} - {\left( {\frac{p}{2}} \right)^2} - 4 &= 0\\ {\left( {x - \frac{p}{2}} \right)^2} - {\left( {\frac{p}{2}} \right)^2} - 4 &= 0\\ {\left( {x - \frac{p}{2}} \right)^2} - \frac{{{p^2}}}{4} - \frac{{16}}{4} &= 0\\ {\left( {x - \frac{p}{2}} \right)^2} - \left( {\frac{{{p^2} + 16}}{4}} \right) &= 0\\ {\left( {x - \frac{p}{2}} \right)^2} &= \left( {\frac{{{p^2} + 16}}{4}} \right)\\ x - \frac{p}{2} &= \pm \sqrt {\frac{{{p^2} + 16}}{4}} \\ x - \frac{p}{2} &= \pm \frac{{\sqrt {{p^2} + 16} }}{{\sqrt 4 }}\\ x - \frac{p}{2} &= \pm \frac{{\sqrt {{p^2} + 16} }}{2}\\ x &= \frac{{p \pm \sqrt {{p^2} + 16} }}{2} \end{align} \)