Aus Online Mathematik Brückenkurs 1

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.