Solution 2.3:4a
From Förberedande kurs i matematik 1
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- | {{ | + | A first thought is perhaps to write the equation as |
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- | {{ | + | {{Displayed math||<math>x^{2}+ax+b=0</math>}} |
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- | < | + | and then try to choose the constants ''a'' and ''b'' in some way so that |
- | {{ | + | <math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation, |
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+ | {{Displayed math||<math>(x+1)(x-2)=0\,\textrm{.}</math>}} | ||
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+ | If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=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 | ||
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+ | {{Displayed math||<math>x^{2}-x-2=0\,\textrm{.}</math>}} | ||
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+ | One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>. | ||
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+ | Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form | ||
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+ | {{Displayed math||<math>ax^{2}-ax-2a=0\,,</math>}} | ||
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+ | where ''a'' is a non-zero constant. |
Current revision
A first thought is perhaps to write the equation as
\displaystyle x^{2}+ax+b=0 |
and then try to choose the constants a and b in some way so that \displaystyle x=-1 and \displaystyle x=2 are solutions. But a better way is to start with a factorized form of a second-order equation,
\displaystyle (x+1)(x-2)=0\,\textrm{.} |
If we consider this equation, we see that both \displaystyle x=-1 and \displaystyle x=2 are solutions to the equation, since \displaystyle x=-1 makes the first factor on the left-hand side zero, whilst \displaystyle x=2 makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
\displaystyle x^{2}-x-2=0\,\textrm{.} |
One answer is thus the equation \displaystyle (x+1)(x-2)=0, or \displaystyle x^{2}-x-2=0\,.
Note: There are actually many answers to this exercise, but what all second-degree equations that have \displaystyle x=-1 and \displaystyle x=2 as roots have in common is that they can be written in the form
\displaystyle ax^{2}-ax-2a=0\,, |
where a is a non-zero constant.