Lösung 2.3:3f

Aus Online Mathematik Brückenkurs 1

Version vom 14:10, 22. Okt. 2008

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)=xx(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

x(x2−2x)+x(2−x)=x2(x−2)−x(x−2)=xx(x−2)−(x−2)=x(x−2)(x−1).

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.