Lösung 2.3:5a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
-
{{Displayed math||<math>(x+7)(x+7)=0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>(x+7)(x+7)=0\,\textrm{.}</math>}}
This equation has only <math>x=-7</math> as a root because both factors become zero only when <math>x=-7</math>. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,
This equation has only <math>x=-7</math> as a root because both factors become zero only when <math>x=-7</math>. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,
-
{{Displayed math||<math>(x+7)(x+7) = x^{2}+14x+49\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>(x+7)(x+7) = x^{2}+14x+49\,\textrm{.}</math>}}
Thus, one answer is the equation <math>x^{2}+14x+49=0\,</math>.
Thus, one answer is the equation <math>x^{2}+14x+49=0\,</math>.
Zeile 12: Zeile 12:
Note: All second-degree equations which have <math>x=-7</math> as its sole root can be written as
Note: All second-degree equations which have <math>x=-7</math> as its sole root can be written as
-
{{Displayed math||<math>ax^{2}+14ax+49a=0\,,</math>}}
+
{{Abgesetzte Formel||<math>ax^{2}+14ax+49a=0\,,</math>}}
where ''a'' is a non-zero constant.
where ''a'' is a non-zero constant.

Version vom 08:33, 22. Okt. 2008

In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation

\displaystyle (x+7)(x+7)=0\,\textrm{.}

This equation has only \displaystyle x=-7 as a root because both factors become zero only when \displaystyle x=-7. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,

\displaystyle (x+7)(x+7) = x^{2}+14x+49\,\textrm{.}

Thus, one answer is the equation \displaystyle x^{2}+14x+49=0\,.


Note: All second-degree equations which have \displaystyle x=-7 as its sole root can be written as

\displaystyle ax^{2}+14ax+49a=0\,,

where a is a non-zero constant.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:5a“