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Aus Online Mathematik Brückenkurs 1

A first thought is perhaps to write the equation as

x2+ax+b=0

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

(x+1)(x−2)=0.

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

x2−x−2=0.

One answer is thus the equation (x+1)(x−2)=0, or x2−x−2=0.

Note: There are actually many answers to this exercise, but what all second-degree equations that have x=−1 and x=2 as roots have in common is that they can be written in the form

ax2−ax−2a=0

where a is a non-zero constant.