Solution 2.3:4a

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m (Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page)
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A first thought is perhaps to write the equation as
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<center> [[Image:2_3_4a-1(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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{{NAVCONTENT_START}}
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<math>x^{2}+ax+b=0</math>
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<center> [[Image:2_3_4a-2(2).gif]] </center>
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and then try to choose the constants
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<math>a</math>
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and
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<math>b</math>
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in some way so that
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<math>x=-\text{1 }</math>
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and
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<math>x=\text{2 }</math>
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are solutions. But a better way is to start with a factorized form of a second-order equation,
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<math>\left( x+1 \right)\left( x-2 \right)=0</math>
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If we consider this equation, we see that both
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<math>x=-\text{1 }</math>
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and
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<math>x=\text{2 }</math>
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are solutions to the equation, since
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<math>x=-\text{1 }</math>
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makes the first factor on the left-hand side zero, whilst
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<math>x=\text{2 }</math>
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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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<math>x^{2}-x-2=0</math>
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 +
 
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One answer is thus the equation
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<math>\left( x+1 \right)\left( x-2 \right)=0</math>, or
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<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
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<math>x=-\text{1 }</math>
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and
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<math>x=\text{2 }</math>
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as roots have in common is that they can be written in the form
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<math>ax^{2}-ax-2a=0</math>
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where
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<math>a</math>
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is a non-zero constant.

Revision as of 09:14, 21 September 2008

A first thought is perhaps to write the equation as


\displaystyle x^{2}+ax+b=0


and then try to choose the constants \displaystyle a and \displaystyle b in some way so that \displaystyle x=-\text{1 } and \displaystyle x=\text{2 } are solutions. But a better way is to start with a factorized form of a second-order equation,


\displaystyle \left( x+1 \right)\left( x-2 \right)=0


If we consider this equation, we see that both \displaystyle x=-\text{1 } and \displaystyle x=\text{2 } are solutions to the equation, since \displaystyle x=-\text{1 } makes the first factor on the left-hand side zero, whilst \displaystyle x=\text{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


One answer is thus the equation \displaystyle \left( x+1 \right)\left( x-2 \right)=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=-\text{1 } and \displaystyle x=\text{2 } as roots have in common is that they can be written in the form


\displaystyle ax^{2}-ax-2a=0


where \displaystyle a is a non-zero constant.

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