Lösung 2.3:5a
Aus Online Mathematik Brückenkurs 1
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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>}} | ||
- | <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, |
- | + | {{Displayed math||<math>(x+7)(x+7) = x^{2}+14x+49\,\textrm{.}</math>}} | |
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- | <math>x= | + | |
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+ | Thus, one answer is the equation <math>x^{2}+14x+49=0\,</math>. | ||
- | <math>\left( x+7 \right)\left( x+7 \right)=x^{2}+14x+49</math> | ||
+ | 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>}} | |
- | <math> | + | |
- | + | where ''a'' is a non-zero constant. | |
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- | where | + | |
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- | is a non-zero constant. | + |
Version vom 10:57, 29. Sep. 2008
In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
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,
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
where a is a non-zero constant.