Aus Online Mathematik Brückenkurs 1
We can split up the first term on the left-hand side, x(x2−2x), into factors by taking x outside the bracket, x(x2−2x)=x
x(x−2) and writing the other term as x(2−x)=−x(x−2). From this we see that both terms contain x(x−2) as common factors and, if we take out those, the left-hand side becomes
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| x(x2−2x)+x(2−x)=x2(x−2)−x(x−2)=x x(x−2)−(x−2) =x(x−2)(x−1). |
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The whole equation can be written as
and this equation is satisfied only when one of the three factors x, x−2 or x−1 is zero, i.e. the solutions are x=0, x=2 and x=1.
Because it is not completely obvious that x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:
- x = 1: LHS=1(12−21)+1(2−1)=1(−1)+11=0=RHS.
