Solution 2.3:4a

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A first thought is perhaps to write the equation as
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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<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
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<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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and then try to choose the constants
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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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<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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{{Displayed math||<math>x^{2}-x-2=0\,\textrm{.}</math>}}
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<math>\left( x+1 \right)\left( x-2 \right)=0</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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If we consider this equation, we see that both
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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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<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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{{Displayed math||<math>ax^{2}-ax-2a=0\,,</math>}}
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<math>x^{2}-x-2=0</math>
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where ''a'' is a non-zero constant.
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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.
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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.

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