Lösung 2.3:4a
Aus Online Mathematik Brückenkurs 1
K (Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page) |
|||
Zeile 1: | Zeile 1: | ||
- | {{ | + | A first thought is perhaps to write the equation as |
- | < | + | |
- | {{ | + | |
- | { | + | <math>x^{2}+ax+b=0</math> |
- | < | + | |
- | {{ | + | |
+ | and then try to choose the constants | ||
+ | <math>a</math> | ||
+ | and | ||
+ | <math>b</math> | ||
+ | in some way so that | ||
+ | <math>x=-\text{1 }</math> | ||
+ | and | ||
+ | <math>x=\text{2 }</math> | ||
+ | are solutions. But a better way is to start with a factorized form of a second-order equation, | ||
+ | |||
+ | |||
+ | <math>\left( x+1 \right)\left( x-2 \right)=0</math> | ||
+ | |||
+ | |||
+ | If we consider this equation, we see that both | ||
+ | <math>x=-\text{1 }</math> | ||
+ | and | ||
+ | <math>x=\text{2 }</math> | ||
+ | are solutions to the equation, since | ||
+ | <math>x=-\text{1 }</math> | ||
+ | makes the first factor on the left-hand side zero, whilst | ||
+ | <math>x=\text{2 }</math> | ||
+ | makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get | ||
+ | |||
+ | |||
+ | <math>x^{2}-x-2=0</math> | ||
+ | |||
+ | |||
+ | One answer is thus the equation | ||
+ | <math>\left( x+1 \right)\left( x-2 \right)=0</math>, or | ||
+ | <math>x^{2}-x-2=0</math>. | ||
+ | |||
+ | NOTE: There are actually many answers to this exercise, but what all second-degree equations that have | ||
+ | <math>x=-\text{1 }</math> | ||
+ | and | ||
+ | <math>x=\text{2 }</math> | ||
+ | as roots have in common is that they can be written in the form | ||
+ | |||
+ | |||
+ | <math>ax^{2}-ax-2a=0</math> | ||
+ | |||
+ | |||
+ | where | ||
+ | <math>a</math> | ||
+ | is a non-zero constant. |
Version vom 09:14, 21. Sep. 2008
A first thought is perhaps to write the equation as
and then try to choose the constants
x+1
x−2
=0
If we consider this equation, we see that both
One answer is thus the equation
x+1
x−2
=0
NOTE: There are actually many answers to this exercise, but what all second-degree equations that have
where